What volume of CO2 gas is generated at 1 atm and 37C from the reaction of 0.350g of NaHCO3 with excess HCl ?
Q. (gastric juice) according to : NaHCO3(s) + HCl(aq) <<>>NaCl(aq)+ H2O(l) + CO2
Asked by tennisgirl28 - Mon Dec 22 21:01:48 2008 - - 1 Answers - 0 Comments
A. Moles NaHCO3 = 0.350 g / 84.0088 g/mol= 0.00417 = moles CO2 T = 37 + 273 = 310 K V = nRT / p = 0.00417 x 0.0821 x 310 / 1 = 0.106 L
Answered by Dr.A - Mon Dec 22 23:50:34 2008
Q. (gastric juice) according to : NaHCO3(s) + HCl(aq) <<>>NaCl(aq)+ H2O(l) + CO2
Asked by tennisgirl28 - Mon Dec 22 21:01:48 2008 - - 1 Answers - 0 Comments
A. Moles NaHCO3 = 0.350 g / 84.0088 g/mol= 0.00417 = moles CO2 T = 37 + 273 = 310 K V = nRT / p = 0.00417 x 0.0821 x 310 / 1 = 0.106 L
Answered by Dr.A - Mon Dec 22 23:50:34 2008
wht is the pH of NaOH, HCl, NaHCO3, Acetic Acid, table salt and soap solution and what are these materials ina?
Q. what is the pH of NaOH, HCl, NaHCO3, Acetic Acid, table salt and soap solution and what are these materials in nature?
Asked by Maryam R - Thu Feb 19 11:59:50 2009 - - 1 Answers - 0 Comments
A. NaOH is used in cleaning products. It has varying pHs depending on its concentration but by itselft it has a pH of 14. HCl is what you find in your stomach and it helps to kill any bacteria on food. pH = 1 Acetic Acid is found in vinegar and had a pH of around 2. Table salt is NaCl (sodium chloride) and has a pH of 7 NaHCO3 is what you use in baking it ot often called bicarbonate of soda/baking soda. it has a pH of 8
Answered by g - Thu Feb 19 12:18:48 2009
Q. what is the pH of NaOH, HCl, NaHCO3, Acetic Acid, table salt and soap solution and what are these materials in nature?
Asked by Maryam R - Thu Feb 19 11:59:50 2009 - - 1 Answers - 0 Comments
A. NaOH is used in cleaning products. It has varying pHs depending on its concentration but by itselft it has a pH of 14. HCl is what you find in your stomach and it helps to kill any bacteria on food. pH = 1 Acetic Acid is found in vinegar and had a pH of around 2. Table salt is NaCl (sodium chloride) and has a pH of 7 NaHCO3 is what you use in baking it ot often called bicarbonate of soda/baking soda. it has a pH of 8
Answered by g - Thu Feb 19 12:18:48 2009
What is the white stuff that appears when adding NaHCO3 to HCl?
Q. I know the product is NaCl + H2CO3, but is the white part the sodium chloride or the carbonic acid. Thanks :)
Asked by sarahshed - Wed Jun 10 16:08:45 2009 - - 3 Answers - 0 Comments
A. We chemists are used to think about precipitates when we are asked about " that white stuff" But what I think you are referring to here is when HCl is added the NaHCO3, strong effervescence, or frothing of the bicarbonate occurs. This looks like a white foam being produced on the top of the reaction. This is the generated CO2 gas foaming up the liquid.
Answered by Trevor H - Wed Jun 10 16:26:21 2009
Q. I know the product is NaCl + H2CO3, but is the white part the sodium chloride or the carbonic acid. Thanks :)
Asked by sarahshed - Wed Jun 10 16:08:45 2009 - - 3 Answers - 0 Comments
A. We chemists are used to think about precipitates when we are asked about " that white stuff" But what I think you are referring to here is when HCl is added the NaHCO3, strong effervescence, or frothing of the bicarbonate occurs. This looks like a white foam being produced on the top of the reaction. This is the generated CO2 gas foaming up the liquid.
Answered by Trevor H - Wed Jun 10 16:26:21 2009
I need to fill a bag with HCl and NaHCO3 without making it pop. How do I do this?
Q. Using stoichiometry, gas laws, molarity, and lab techniques, you and your lab partner will fully inflate, but not burst, a sip-lock bag (holds about 4 L). The reactants you will be using are sodium hydrogen carbonate and hydrochloric acid. There should be no reactant in excess. How do I do this?
Asked by tnsstr - Wed Mar 11 15:43:20 2009 - - 1 Answers - 0 Comments
A. I'm assuming you're doing this on paper and not in real life since you would run into issues with the concentration of HCl and using a thin plastic bag which would melt from the heat produced. Your goal is to calculate the volume of the liquid HCl and the volume of the powdered NaHCO3 and how much gas volume (CO2) it will produce when mixed, so that the volume of gas, liquid and powder does not exceed a total of 4.0 Litres. You should be able to assume no gas exists within the bag when you start the process. So, you'll start off with this... VHcl + VNaHCO3 = X (Total Volume of the Hcl & Sodium Bicarb) 4.0Litres - X = Y (Volume of gas produced) Now you'll need to figure out how much gas is produced by any volume of Hcl + NaHCO3… [cont.]
Answered by Gary D - Wed Mar 11 15:55:48 2009
Q. Using stoichiometry, gas laws, molarity, and lab techniques, you and your lab partner will fully inflate, but not burst, a sip-lock bag (holds about 4 L). The reactants you will be using are sodium hydrogen carbonate and hydrochloric acid. There should be no reactant in excess. How do I do this?
Asked by tnsstr - Wed Mar 11 15:43:20 2009 - - 1 Answers - 0 Comments
A. I'm assuming you're doing this on paper and not in real life since you would run into issues with the concentration of HCl and using a thin plastic bag which would melt from the heat produced. Your goal is to calculate the volume of the liquid HCl and the volume of the powdered NaHCO3 and how much gas volume (CO2) it will produce when mixed, so that the volume of gas, liquid and powder does not exceed a total of 4.0 Litres. You should be able to assume no gas exists within the bag when you start the process. So, you'll start off with this... VHcl + VNaHCO3 = X (Total Volume of the Hcl & Sodium Bicarb) 4.0Litres - X = Y (Volume of gas produced) Now you'll need to figure out how much gas is produced by any volume of Hcl + NaHCO3… [cont.]
Answered by Gary D - Wed Mar 11 15:55:48 2009
is the reaction between NaHCO3 and HCl exothermic or endothermic?
Q. is the reaction between NaHCO3 and HCl exothermic or endothermic?
Asked by Dude123 - Tue Apr 8 00:21:23 2008 - - 6 Answers - 0 Comments
A. Offhand, I'm pretty sure that this reaction is exothermic. You could always figure it out using Hess's law. DH = sum(DHf products) - sum(DHf reactants) NaHCO3 + HCl(aq)--> NaCl(aq) + CO2(g) + H2O(l) The is the net ionic equation. These are the only ions and compounds for which you need to find the heats of formation (DHf). HCO3- + H+ --> CO2(g) + H2O(l) === By the way, just because a reaction is spontaneous doesn't make it exothermic. Spontaneity is given by Gibbs free energy. DG = DH - TDS.
Answered by pisgahchemist - Tue Apr 8 00:35:14 2008
Q. is the reaction between NaHCO3 and HCl exothermic or endothermic?
Asked by Dude123 - Tue Apr 8 00:21:23 2008 - - 6 Answers - 0 Comments
A. Offhand, I'm pretty sure that this reaction is exothermic. You could always figure it out using Hess's law. DH = sum(DHf products) - sum(DHf reactants) NaHCO3 + HCl(aq)--> NaCl(aq) + CO2(g) + H2O(l) The is the net ionic equation. These are the only ions and compounds for which you need to find the heats of formation (DHf). HCO3- + H+ --> CO2(g) + H2O(l) === By the way, just because a reaction is spontaneous doesn't make it exothermic. Spontaneity is given by Gibbs free energy. DG = DH - TDS.
Answered by pisgahchemist - Tue Apr 8 00:35:14 2008
Reaction between NaHCO3 + HCl?
Q. Reaction between NaHCO3 + HCl, The chemical equation is NaHCO3 + HCl --> NaCl + H2O + CO2 The question is What form do these products take in the solution? (hint are they ions or compounds?)
Asked by on_gt - Thu Mar 15 10:13:11 2007 - - 3 Answers - 0 Comments
A. NaHCO3(aq) + HCl(aq) --> NaCl(aq) + H2O(l) + CO2(g) CO2 comes bubbling out, NaCl is a dissolved salt water is - well - just water
Answered by Dr Dave P - Thu Mar 15 10:18:49 2007
Q. Reaction between NaHCO3 + HCl, The chemical equation is NaHCO3 + HCl --> NaCl + H2O + CO2 The question is What form do these products take in the solution? (hint are they ions or compounds?)
Asked by on_gt - Thu Mar 15 10:13:11 2007 - - 3 Answers - 0 Comments
A. NaHCO3(aq) + HCl(aq) --> NaCl(aq) + H2O(l) + CO2(g) CO2 comes bubbling out, NaCl is a dissolved salt water is - well - just water
Answered by Dr Dave P - Thu Mar 15 10:18:49 2007
NaHCO3 + HCl = NaCl + H2O + CO2 spectator ions?
Q. i have the balanced equation NaHCO3(s) + HCl = NaCl + H2O + CO2 NaHCO3 is a solid...i need to know the net ionic equation. i came up with: HCO3[-] + H[+] = H2O + CO2 are you able to split NaHCO3(s) into Na[+] and HCO3[+] because it is a solid? is my net ionic equation correct? i think it's either that or: NaHCO3 + H[+] = Na[+] + H2O + CO2 any help would be appreciated. thanks
Asked by George P - Sun Feb 24 12:08:18 2008 - - 2 Answers - 2 Comments
A. Solids are never split into ions. So the molecular equation is NaHCO3(s) + H[+] + Cl[-] = Na[+] + Cl[-] + H2O + CO2 Cl[-] is a spectator ion. NaHCO3(s) + H[+] = Na[+] + H2O + CO2 Your second equation is correct. Only Cl[-] is a spectator ion.
Answered by Some Body - Sun Feb 24 12:17:31 2008
Q. i have the balanced equation NaHCO3(s) + HCl = NaCl + H2O + CO2 NaHCO3 is a solid...i need to know the net ionic equation. i came up with: HCO3[-] + H[+] = H2O + CO2 are you able to split NaHCO3(s) into Na[+] and HCO3[+] because it is a solid? is my net ionic equation correct? i think it's either that or: NaHCO3 + H[+] = Na[+] + H2O + CO2 any help would be appreciated. thanks
Asked by George P - Sun Feb 24 12:08:18 2008 - - 2 Answers - 2 Comments
A. Solids are never split into ions. So the molecular equation is NaHCO3(s) + H[+] + Cl[-] = Na[+] + Cl[-] + H2O + CO2 Cl[-] is a spectator ion. NaHCO3(s) + H[+] = Na[+] + H2O + CO2 Your second equation is correct. Only Cl[-] is a spectator ion.
Answered by Some Body - Sun Feb 24 12:17:31 2008
NaHCO3 + HCL --> H2O + NaCl + CO2 Help...?
Q. NaHCO3 + HCL --> H2O + NaCl + CO2 Using that equation (which is already balanced), I'm supposed to find out how many moles of HCL "would have been used up" in the reaction...? Does anyone know how to do this? Thanks, I got the answer
Asked by Igg - Sun Jan 6 16:07:33 2008 - - 4 Answers - 0 Comments
A. The same number of the moles of NaHCO3
Answered by Dr.A - Sun Jan 6 16:10:57 2008
Q. NaHCO3 + HCL --> H2O + NaCl + CO2 Using that equation (which is already balanced), I'm supposed to find out how many moles of HCL "would have been used up" in the reaction...? Does anyone know how to do this? Thanks, I got the answer
Asked by Igg - Sun Jan 6 16:07:33 2008 - - 4 Answers - 0 Comments
A. The same number of the moles of NaHCO3
Answered by Dr.A - Sun Jan 6 16:10:57 2008
Calculate the grams of salt expected from the reaction: NaHCO3 + HCl -> NaCl + H2O + CO2 (g)?
Q. you have to do the theoretical yield but im not sure if you have only find the moles of Na or if you would find the moles of the NaHCO3... The starting weight of the NaHCO3 is 2.01 grams.
Asked by em - Thu May 22 19:37:39 2008 - - 2 Answers - 0 Comments
A. mol wt of NaHCO3 = 23 + 1+12+48 = 84g NaCl =23 + 35.5 = 58.5g 84g 0f NaHCO3 produces 58.5g of NaCl 2.01g of NaHCO3 will produce 1.399821 g of NaCl
Answered by unknown - Sat May 24 00:51:48 2008
Q. you have to do the theoretical yield but im not sure if you have only find the moles of Na or if you would find the moles of the NaHCO3... The starting weight of the NaHCO3 is 2.01 grams.
Asked by em - Thu May 22 19:37:39 2008 - - 2 Answers - 0 Comments
A. mol wt of NaHCO3 = 23 + 1+12+48 = 84g NaCl =23 + 35.5 = 58.5g 84g 0f NaHCO3 produces 58.5g of NaCl 2.01g of NaHCO3 will produce 1.399821 g of NaCl
Answered by unknown - Sat May 24 00:51:48 2008
What is the balanced equation of HCl+NaHCO3->NaCl+CO2+H 2O?
Q. What is the balanced equation of HCl+NaHCO3->NaCl+CO2+H 2O?
Asked by Natalie - Wed Jul 29 20:30:13 2009 - - 1 Answers - 0 Comments
Q. What is the balanced equation of HCl+NaHCO3->NaCl+CO2+H 2O?
Asked by Natalie - Wed Jul 29 20:30:13 2009 - - 1 Answers - 0 Comments
HCl + NaHCO3 -- Experimental Formula?
Q. When you add Hydrocloric Acid to Sodium Bicarbonate HCl + NaHCO3 ---> NaCl + HHCO3 is this the outcome? and, do you just put the two H's beside each other in HHCO3?
Asked by Smyxz - Tue Mar 31 00:33:42 2009 - - 1 Answers - 0 Comments
A. outcome should be: HCl + NaHCO3 ---> NaCl + CO2 + H2O
Answered by Erica L - Tue Mar 31 00:38:34 2009
Q. When you add Hydrocloric Acid to Sodium Bicarbonate HCl + NaHCO3 ---> NaCl + HHCO3 is this the outcome? and, do you just put the two H's beside each other in HHCO3?
Asked by Smyxz - Tue Mar 31 00:33:42 2009 - - 1 Answers - 0 Comments
A. outcome should be: HCl + NaHCO3 ---> NaCl + CO2 + H2O
Answered by Erica L - Tue Mar 31 00:38:34 2009
Whats the balanced equation for NaHCO3 + HCl?
Q. Its for a lab that I'm doing and I don't know how to do it. I know that one of the products will be CO2 but I don't know the rest. I tried to do it myself but failed. Thank you! Thank you all for helping me out and giving me the right answer but unfortunantly i have to pick one best answer.
Asked by Jesus D. - Thu Jul 12 00:08:55 2007 - - 5 Answers - 0 Comments
A. CO2 NaCl H2O Seems right to me, but I'm no chemist. Base + Acid typically yields some type of salt. In this case, also CO2 gaseous. Usually, you'll see HCl + NaOH = NaCl + H2O (Hydrocloric Acid, and Lye = Salt water, and a LOT of heat.)
Answered by A Military Veteran - Thu Jul 12 00:15:58 2007
Q. Its for a lab that I'm doing and I don't know how to do it. I know that one of the products will be CO2 but I don't know the rest. I tried to do it myself but failed. Thank you! Thank you all for helping me out and giving me the right answer but unfortunantly i have to pick one best answer.
Asked by Jesus D. - Thu Jul 12 00:08:55 2007 - - 5 Answers - 0 Comments
A. CO2 NaCl H2O Seems right to me, but I'm no chemist. Base + Acid typically yields some type of salt. In this case, also CO2 gaseous. Usually, you'll see HCl + NaOH = NaCl + H2O (Hydrocloric Acid, and Lye = Salt water, and a LOT of heat.)
Answered by A Military Veteran - Thu Jul 12 00:15:58 2007
How many mole of salt (NaCl) are expect for every one mole of NaHCO3 that reacts?
Q. NaHCO3 + HCl -> NaCl + H2O + CO2 (g) is the equation and you start with 2.01 grams of NaHCO3. So if the next part of the question is write the mole ratio for the moles of NaCl over the moles of NaHCO3, then would it just be 0.0239 moles over 0.0239 moles?
Asked by em - Thu May 22 20:06:52 2008 - - 2 Answers - 0 Comments
A. This is a simple and balanced equation. The equation itself says that one mole of NaHCO3 will give one mole of NaCl. So 2.01 gms = 0.0239 Mole of NaHCO3 will give you 0.0239 moles of Na Cl which is 1.3998 gms ~ 9Almost equal to) 1.4 gms.
Answered by Abhijit Purohit - Sun May 25 01:52:12 2008
Q. NaHCO3 + HCl -> NaCl + H2O + CO2 (g) is the equation and you start with 2.01 grams of NaHCO3. So if the next part of the question is write the mole ratio for the moles of NaCl over the moles of NaHCO3, then would it just be 0.0239 moles over 0.0239 moles?
Asked by em - Thu May 22 20:06:52 2008 - - 2 Answers - 0 Comments
A. This is a simple and balanced equation. The equation itself says that one mole of NaHCO3 will give one mole of NaCl. So 2.01 gms = 0.0239 Mole of NaHCO3 will give you 0.0239 moles of Na Cl which is 1.3998 gms ~ 9Almost equal to) 1.4 gms.
Answered by Abhijit Purohit - Sun May 25 01:52:12 2008
NaHCO3 Buffer question & how much HCl to add to make pH7?
Q. carbonic acid, H2CO3, has a pKa1 = 6.37 and a pKa2 = 10.33. began with 7.5 mL of 0.41 M NaHCO3, how many mL of HCl do I need to add to get a pH 7 buffer?
Asked by A A - Thu Oct 25 18:27:23 2007 - - 1 Answers - 0 Comments
A. NaHCO3 is a mono acidic salt, and HCl is mono basic acid. Now NaHCO3 + HCL ---> NaCl + CO2 + H2O So one mole of NaHCO3 will require one mole of HCl. In this case you will need 7.5 ml of 0.41 m HCl, or you can calculate it from the most common equation M1 * V1 = M2 * V2 where M1 and M2 are molarity of acid and base respectively and V1 and V2 are volumes of acid and base respectively . One important point here is CO2. Presence of CO2 will indicate wrong pH while neutralisation. Any CO2 dissolved in water will form carbonic acid and will indicate pH in acidic range. When all CO2 will escape from the solution it will indicate correct pH - neutral
Answered by Abhijit Purohit - Mon Oct 29 07:20:18 2007
Q. carbonic acid, H2CO3, has a pKa1 = 6.37 and a pKa2 = 10.33. began with 7.5 mL of 0.41 M NaHCO3, how many mL of HCl do I need to add to get a pH 7 buffer?
Asked by A A - Thu Oct 25 18:27:23 2007 - - 1 Answers - 0 Comments
A. NaHCO3 is a mono acidic salt, and HCl is mono basic acid. Now NaHCO3 + HCL ---> NaCl + CO2 + H2O So one mole of NaHCO3 will require one mole of HCl. In this case you will need 7.5 ml of 0.41 m HCl, or you can calculate it from the most common equation M1 * V1 = M2 * V2 where M1 and M2 are molarity of acid and base respectively and V1 and V2 are volumes of acid and base respectively . One important point here is CO2. Presence of CO2 will indicate wrong pH while neutralisation. Any CO2 dissolved in water will form carbonic acid and will indicate pH in acidic range. When all CO2 will escape from the solution it will indicate correct pH - neutral
Answered by Abhijit Purohit - Mon Oct 29 07:20:18 2007
what kind of reaction is NaHCO3 + HCl ==> NaCl + H20 + CO2?
Q. what kind of reaction is NaHCO3 + HCl ==> NaCl + H20 + CO2?
Asked by Kyle - Tue Jan 27 19:45:30 2009 - - 2 Answers - 0 Comments
A. it is an acid on carbonate reaction.
Answered by Andy - Tue Jan 27 19:48:58 2009
Q. what kind of reaction is NaHCO3 + HCl ==> NaCl + H20 + CO2?
Asked by Kyle - Tue Jan 27 19:45:30 2009 - - 2 Answers - 0 Comments
A. it is an acid on carbonate reaction.
Answered by Andy - Tue Jan 27 19:48:58 2009
.5 g of NaHCO3 treated with 40 ml HCL. Excess HCL back titrated with 9.5 ml Of .31 M NaOh. Wht is M of HCL?
Q. given mmol HCL = mmol NAHCO3 + mmol Naoh
Asked by om - Tue May 5 12:19:52 2009 - - 2 Answers - 0 Comments
Q. given mmol HCL = mmol NAHCO3 + mmol Naoh
Asked by om - Tue May 5 12:19:52 2009 - - 2 Answers - 0 Comments
Is the neutralization of a strong acid, for example, HCL, with a weak base, for example NaHCO3, the solution ?
Q. Is the neutralization of a strong acid, for example, HCL, with a weak base, for example NaHCO3, the solution ? The solution that is produced is A. neutral B. slightly acidic C. Slightly base D. could be acidic or base 2. The [H3O ] in a solution is 1 X 10^-9 M A. pOH = 5 B. [OH-] < 1 X 10 ^-7 M C. the solution is acidic d. all the above
Asked by cbvet85 - Thu Jul 31 09:53:40 2008 - - 4 Answers - 0 Comments
A. TO neutralize a strong acid, you need a strong base. When you try to neutralize a strong acid with a weak base you are left with a solution that is still slightly acidic. It all has to do with the amount of H+ that are exchanged. So, the answer is B...slightly acidic For 2...the solution is acidic
Answered by Discombobulated - Thu Jul 31 10:30:18 2008
Q. Is the neutralization of a strong acid, for example, HCL, with a weak base, for example NaHCO3, the solution ? The solution that is produced is A. neutral B. slightly acidic C. Slightly base D. could be acidic or base 2. The [H3O ] in a solution is 1 X 10^-9 M A. pOH = 5 B. [OH-] < 1 X 10 ^-7 M C. the solution is acidic d. all the above
Asked by cbvet85 - Thu Jul 31 09:53:40 2008 - - 4 Answers - 0 Comments
A. TO neutralize a strong acid, you need a strong base. When you try to neutralize a strong acid with a weak base you are left with a solution that is still slightly acidic. It all has to do with the amount of H+ that are exchanged. So, the answer is B...slightly acidic For 2...the solution is acidic
Answered by Discombobulated - Thu Jul 31 10:30:18 2008
How would you go about making a standard solution of NaHCO3?
Q. I have been given the task of making a standard solution of NaHCO3 with 5g of its solid, and one that will neutralise with 25cm^3 of HCl. I also need to calculate the exact amounts, so i cant put 'use excess HCl'. The HCl is readily available at whatever concentration is required, to neutralise the NaHCO3. And i need to know the concentration of both the HCl and NaHCO3. I know its quite alot to ask, and i am sorry for this, its just I really need some help. Could you put all your calculation steps in, to help me understand. Many Thanks. :) Sorry, I probably should have added that it is a calorimetry experiment trying to find the enthalpy for the reaction. ( i can do that bit though...i hope)
Asked by AAA - Wed Mar 14 13:45:59 2007 - - 1 Answers - 0 Comments
A. So you want to make a standard solution of NaHCO3 with 5 g NaHCO3 that will neutralize 25 ml of some concentration of HCl solution. First, lets calculate what the M of HCl would be in this case 5 g NaHCO3 * mole/84g = 0.059 moles 0.25L * x = 0.059 mole ==> x = 0.238M HCl Since this is such a high concentration and you are going to be using 5 g NaHCO3 each time it is convenient that you weight out a large amount 25.00g and using a 500 ml volumetic flash add the 25.00 grams to it and add water up to the mark. your final NaHCO3 concentration will be 25g(mole/84g)/.5L = .59 M NaHCO3 Then pipet out 100 mls of solution and titrate is with the 0.238 M HCl 0.1L(0.59M NaHCO3) = 0.059 moles NaHCO3 = 0.25L(0.238MHCl) = 0.059 moles HCl
Answered by Dr Dave P - Wed Mar 14 14:13:53 2007
Q. I have been given the task of making a standard solution of NaHCO3 with 5g of its solid, and one that will neutralise with 25cm^3 of HCl. I also need to calculate the exact amounts, so i cant put 'use excess HCl'. The HCl is readily available at whatever concentration is required, to neutralise the NaHCO3. And i need to know the concentration of both the HCl and NaHCO3. I know its quite alot to ask, and i am sorry for this, its just I really need some help. Could you put all your calculation steps in, to help me understand. Many Thanks. :) Sorry, I probably should have added that it is a calorimetry experiment trying to find the enthalpy for the reaction. ( i can do that bit though...i hope)
Asked by AAA - Wed Mar 14 13:45:59 2007 - - 1 Answers - 0 Comments
A. So you want to make a standard solution of NaHCO3 with 5 g NaHCO3 that will neutralize 25 ml of some concentration of HCl solution. First, lets calculate what the M of HCl would be in this case 5 g NaHCO3 * mole/84g = 0.059 moles 0.25L * x = 0.059 mole ==> x = 0.238M HCl Since this is such a high concentration and you are going to be using 5 g NaHCO3 each time it is convenient that you weight out a large amount 25.00g and using a 500 ml volumetic flash add the 25.00 grams to it and add water up to the mark. your final NaHCO3 concentration will be 25g(mole/84g)/.5L = .59 M NaHCO3 Then pipet out 100 mls of solution and titrate is with the 0.238 M HCl 0.1L(0.59M NaHCO3) = 0.059 moles NaHCO3 = 0.25L(0.238MHCl) = 0.059 moles HCl
Answered by Dr Dave P - Wed Mar 14 14:13:53 2007
NaHCO3(s) + HCl (aq) --> NaCl (aq) + CO2(g) +H2O(l)?
Q. if 100mL of 6 M HCl is spilled on a lab bench, what is the minimum mass of sodium bicarbonate that must be used to neutralize the acid? need helppp
Asked by annie - Fri Nov 2 20:30:24 2007 - - 1 Answers - 0 Comments
A. Moles HCl = 100 x 6 / 1000 = 0.6 the ratio between NaHCO3 and HCl is 1:1 so we need 0.6 mole NaHCO3 ( MM = 84 g/mol ) ==>50.4 g
Answered by Dr.A - Sat Nov 3 01:33:08 2007
Q. if 100mL of 6 M HCl is spilled on a lab bench, what is the minimum mass of sodium bicarbonate that must be used to neutralize the acid? need helppp
Asked by annie - Fri Nov 2 20:30:24 2007 - - 1 Answers - 0 Comments
A. Moles HCl = 100 x 6 / 1000 = 0.6 the ratio between NaHCO3 and HCl is 1:1 so we need 0.6 mole NaHCO3 ( MM = 84 g/mol ) ==>50.4 g
Answered by Dr.A - Sat Nov 3 01:33:08 2007
: HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g) homo or heterogenus mixtures and solutions?
Q. : HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g) homo or heterogenus mixtures and solutions?
Asked by ay bay bay - Fri Nov 23 19:50:23 2007 - - 1 Answers - 0 Comments
A. solutions are homogeneous mixtures
Answered by science teacher - Fri Nov 23 20:26:27 2007
Q. : HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g) homo or heterogenus mixtures and solutions?
Asked by ay bay bay - Fri Nov 23 19:50:23 2007 - - 1 Answers - 0 Comments
A. solutions are homogeneous mixtures
Answered by science teacher - Fri Nov 23 20:26:27 2007
From Yahoo Answer Search: 'NaHCO3 HCl'
Tue Nov 10 22:37:17 2009 [ refresh local cache ]
