When excess silver nitrate was added to a 25.0 mL sample of a solution of calcium chloride?
Q. When excess silver nitrate was added to a 25.0 mL sample of a solution of calcium chloride, 0.9256 gram of silver chloride precipitated. What is the concentration of the calcium chloride solution? can you guys show me the work?
Asked by Faggg - Sat Feb 13 20:37:13 2010 - - 1 Answers - 0 Comments
A. use molar mass to find moles 0.9256 g AgCl @ 143.32 g/mol = 0.006458 moles of AgCl by the reaction 2 AgNO3 & 1 CaCL2 --> 1 Ca(NO3)2 & 2 AgCl 0.006458 moles of AgCl is produced from 1/2 as many moles of CaCl2 = 0.003229 moles CaCl2 find molarity 0.003229 moles CaCl2 / 0.0250 litres = 0.1292 Molar your answer rounded to 3 sig figs 0.129 Molar CaCl2
Answered by Steve O - Sat Feb 13 21:39:00 2010
Q. When excess silver nitrate was added to a 25.0 mL sample of a solution of calcium chloride, 0.9256 gram of silver chloride precipitated. What is the concentration of the calcium chloride solution? can you guys show me the work?
Asked by Faggg - Sat Feb 13 20:37:13 2010 - - 1 Answers - 0 Comments
A. use molar mass to find moles 0.9256 g AgCl @ 143.32 g/mol = 0.006458 moles of AgCl by the reaction 2 AgNO3 & 1 CaCL2 --> 1 Ca(NO3)2 & 2 AgCl 0.006458 moles of AgCl is produced from 1/2 as many moles of CaCl2 = 0.003229 moles CaCl2 find molarity 0.003229 moles CaCl2 / 0.0250 litres = 0.1292 Molar your answer rounded to 3 sig figs 0.129 Molar CaCl2
Answered by Steve O - Sat Feb 13 21:39:00 2010
6 liters of hydrochloric acid reacts with excess silver nitrate. How much silver chloride will be produced?
Q. answer in grams
Asked by ai - Tue Jan 15 20:00:08 2008 - - 3 Answers - 0 Comments
A. We need to know the concentration of the hydrochloric acid first. Since you are using 6 liters I would assume that you are going to use the commercial grade of concentrated hydrochloric acid which is ~12 molar Assuming that it is 12 molar you have the equation that mols of HCL = mols of silver nitrate = mol of silver choride HCl(aq)+ AgNos(aq) = AgCl(s) + HNO3 (aq) mols of HCl = M x V = 12.0 x 6.0liter = 72 mols 1mol of silver chloride = 142 g so 72 x142 = 10256g or 10.2 Kg of AgCl That is a lot of silver chloride!!!
Answered by George F - Tue Jan 15 20:18:52 2008
Q. answer in grams
Asked by ai - Tue Jan 15 20:00:08 2008 - - 3 Answers - 0 Comments
A. We need to know the concentration of the hydrochloric acid first. Since you are using 6 liters I would assume that you are going to use the commercial grade of concentrated hydrochloric acid which is ~12 molar Assuming that it is 12 molar you have the equation that mols of HCL = mols of silver nitrate = mol of silver choride HCl(aq)+ AgNos(aq) = AgCl(s) + HNO3 (aq) mols of HCl = M x V = 12.0 x 6.0liter = 72 mols 1mol of silver chloride = 142 g so 72 x142 = 10256g or 10.2 Kg of AgCl That is a lot of silver chloride!!!
Answered by George F - Tue Jan 15 20:18:52 2008
A 5.00 g piece of copper is placed in a solution of silver nitrate (AgNO3) in which there is excess silver nit?
Q. The product produced in this reaction is copper nitrate, Cu(NO3)2. The mass of the silver produced is 15.2g. What is the percent yield for this reaction? Could someone PLEASE help me with this chemistry problem? Im completely stumped.
Asked by Kris - Tue Sep 16 18:55:32 2008 - - 2 Answers - 0 Comments
A. First you need a balanced chem equation for the reaction. This would be: Cu + 2Ag(NO3) --> 2Ag + Cu(NO3)2. Therefore we know that for every two moles of silver formed we need one mole of copper. We are starting with 5g of copper, whose molar mass is 63.6 g/mol, therefore we have 5/63.6 = 0.079 mol copper. If all of the moles of copper were converted to silver we would get 2x0.079 mol silver (because of the 1:2 relationship in the equation) which equals 0.158 moles of silver. Silver has a molar mass of 107.9g/mol, so if we did get 0.158 mol of silver, this would be 107.9 x 0.158 = 17.0g silver. Because we only got 15.2 g, we know that the extent of reaction was less than 100%. The actual percent yield for the reaction is the actual amount… [cont.]
Answered by ChemJes - Tue Sep 16 19:07:14 2008
Q. The product produced in this reaction is copper nitrate, Cu(NO3)2. The mass of the silver produced is 15.2g. What is the percent yield for this reaction? Could someone PLEASE help me with this chemistry problem? Im completely stumped.
Asked by Kris - Tue Sep 16 18:55:32 2008 - - 2 Answers - 0 Comments
A. First you need a balanced chem equation for the reaction. This would be: Cu + 2Ag(NO3) --> 2Ag + Cu(NO3)2. Therefore we know that for every two moles of silver formed we need one mole of copper. We are starting with 5g of copper, whose molar mass is 63.6 g/mol, therefore we have 5/63.6 = 0.079 mol copper. If all of the moles of copper were converted to silver we would get 2x0.079 mol silver (because of the 1:2 relationship in the equation) which equals 0.158 moles of silver. Silver has a molar mass of 107.9g/mol, so if we did get 0.158 mol of silver, this would be 107.9 x 0.158 = 17.0g silver. Because we only got 15.2 g, we know that the extent of reaction was less than 100%. The actual percent yield for the reaction is the actual amount… [cont.]
Answered by ChemJes - Tue Sep 16 19:07:14 2008
1. 6 L of hydrochloric acid reacts w/ excess silver nitrate, how much silver chloride in g will be produced.?
Q. If possible can you answer the question using dimensional analysis.
Asked by wealthyxhobo - Wed Jan 16 19:41:26 2008 - - 1 Answers - 0 Comments
A. We must know the molarity of HCl Moles = molarity x 1.6 => moles of AgCl Multiply by molar mass of AgCl
Answered by Dr.A - Thu Jan 17 05:09:44 2008
Q. If possible can you answer the question using dimensional analysis.
Asked by wealthyxhobo - Wed Jan 16 19:41:26 2008 - - 1 Answers - 0 Comments
A. We must know the molarity of HCl Moles = molarity x 1.6 => moles of AgCl Multiply by molar mass of AgCl
Answered by Dr.A - Thu Jan 17 05:09:44 2008
A .256g sample of CoCL3 *XH20 was dissolved and excess silver nitrate was added. Dried and filtered it weighed?
Q. A .256g sample of CoCL3 *XH20 was dissolved and excess silver nitrate was added. Dried and filtered it weighed?
Asked by flaca_1121us - Fri Oct 17 09:55:16 2008 - - 1 Answers - 0 Comments
A. and the question is ?? incomplete question, do u wanna know the x of the hydrated form, or what??
Answered by kitty_23230 - Fri Oct 17 10:01:33 2008
Q. A .256g sample of CoCL3 *XH20 was dissolved and excess silver nitrate was added. Dried and filtered it weighed?
Asked by flaca_1121us - Fri Oct 17 09:55:16 2008 - - 1 Answers - 0 Comments
A. and the question is ?? incomplete question, do u wanna know the x of the hydrated form, or what??
Answered by kitty_23230 - Fri Oct 17 10:01:33 2008
If 21.5 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.855 g of preci?
Q. If 21.5 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.855 g of precipitate, what is the molarity of silver ion in the original solution?
Asked by Ade S - Wed Oct 1 00:32:46 2008 - - 1 Answers - 0 Comments
A. Moles AgCl = 0.855 g/ 143.32 g/mol =0.00597 [Ag+] = 0.00597 / 0.0215 L =0.277 M
Answered by Dr.A - Wed Oct 1 10:10:27 2008
Q. If 21.5 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.855 g of precipitate, what is the molarity of silver ion in the original solution?
Asked by Ade S - Wed Oct 1 00:32:46 2008 - - 1 Answers - 0 Comments
A. Moles AgCl = 0.855 g/ 143.32 g/mol =0.00597 [Ag+] = 0.00597 / 0.0215 L =0.277 M
Answered by Dr.A - Wed Oct 1 10:10:27 2008
We add excess Na2CrO4 solution to 43.0 mL of a solution of silver nitrate (AgNO3) to form insoluble solid Ag2C?
Q. We add excess Na2CrO4 solution to 43.0 mL of a solution of silver nitrate (AgNO3) to form insoluble solid Ag2CrO4. When it has been dried and weighed, the mass of Ag2CrO4 is found to be 0.570 grams. What is the molarity of the AgNO3 solution? Answer in units of M.
Asked by Millie - Mon Nov 9 12:32:48 2009 - - 1 Answers - 0 Comments
A. Molar mass Ag2CrO4 = 331.73 g/mol 2 AgNO3 + Na2CrO4 = Ag2CrO4 + 2 NaNO3 moles Ag2CrO4 = 0.570 g/ 331.73 g/mol=0.00172 the ratio between AgNO3 and Ag2CrO4 is 2 : 1 moles AgNO3 = 2 x 0.00172 =0.00344 M = 0.00344 / 0.0430 L=0.0800
Answered by Dr.A - Mon Nov 9 12:46:31 2009
Q. We add excess Na2CrO4 solution to 43.0 mL of a solution of silver nitrate (AgNO3) to form insoluble solid Ag2CrO4. When it has been dried and weighed, the mass of Ag2CrO4 is found to be 0.570 grams. What is the molarity of the AgNO3 solution? Answer in units of M.
Asked by Millie - Mon Nov 9 12:32:48 2009 - - 1 Answers - 0 Comments
A. Molar mass Ag2CrO4 = 331.73 g/mol 2 AgNO3 + Na2CrO4 = Ag2CrO4 + 2 NaNO3 moles Ag2CrO4 = 0.570 g/ 331.73 g/mol=0.00172 the ratio between AgNO3 and Ag2CrO4 is 2 : 1 moles AgNO3 = 2 x 0.00172 =0.00344 M = 0.00344 / 0.0430 L=0.0800
Answered by Dr.A - Mon Nov 9 12:46:31 2009
A 10.0mL sample of potassium iodide solution was analyzed by adding an excess of silver nitrate solution to..?
Q. produce silver iodide crystals, which were filtered from the solution KI + AgNO3-> KNO3 + AgI If 2.183g of silver iodide was obtained, what was the molarity of the orginal KI solution?
Asked by blahblahblah - Fri Oct 17 00:23:26 2008 - - 2 Answers - 0 Comments
Q. produce silver iodide crystals, which were filtered from the solution KI + AgNO3-> KNO3 + AgI If 2.183g of silver iodide was obtained, what was the molarity of the orginal KI solution?
Asked by blahblahblah - Fri Oct 17 00:23:26 2008 - - 2 Answers - 0 Comments
have a grey stain on my skin from silver nitrate after removing excess skin will it go away and how long?
Q. what to know how long it will take to go away.
Asked by attwood - Fri Mar 16 02:56:00 2007 - - 1 Answers - 0 Comments
A. It should last about a week or so, and will go away by itself. If you want, you can try a salt water solution to try to help it fade?
Answered by Jeffrey C - Fri Mar 16 03:04:22 2007
Q. what to know how long it will take to go away.
Asked by attwood - Fri Mar 16 02:56:00 2007 - - 1 Answers - 0 Comments
A. It should last about a week or so, and will go away by itself. If you want, you can try a salt water solution to try to help it fade?
Answered by Jeffrey C - Fri Mar 16 03:04:22 2007
when 21.8g of silver nitrate is reacted with excess sodium chloride, 17.8 g of silver chloride is formed, what?
Q. is the precent yield of silver chloride?
Asked by rk - Tue Jan 6 01:25:31 2009 - - 1 Answers - 0 Comments
A. First, write the balanced equation. AgNO3 + NaCl --> AgCl + NaNO3 According to the question, AgNO3 is the limiting reagent (excess NaCl means there's more than you need). That means that all of it will be consumed in the reaction, so however much you have determines how much silver chloride can be formed. Convert g AgNO3 to mol AgNO3 to mol AgCl to g AgCl, and that will be your theoretical yield. Your actual yield is 17.8 g, and so if you divide actual yield/theoretical yield * 100 you get your percent yield. You need molecular weights for this: MW AgNO3 = 170 g/mol MW AgCl = 143 g/mol theoretical yield: 21.8 g AgNO3 * (1 mol AgNO3 / 170 g AgNO3) * (1 mol AgCl / 1 mol AgNO3) * (143 g AuCl / mol AuCl) = 18.3 g Percent yield: 17.8… [cont.]
Answered by Chemis-T - Tue Jan 6 02:40:35 2009
Q. is the precent yield of silver chloride?
Asked by rk - Tue Jan 6 01:25:31 2009 - - 1 Answers - 0 Comments
A. First, write the balanced equation. AgNO3 + NaCl --> AgCl + NaNO3 According to the question, AgNO3 is the limiting reagent (excess NaCl means there's more than you need). That means that all of it will be consumed in the reaction, so however much you have determines how much silver chloride can be formed. Convert g AgNO3 to mol AgNO3 to mol AgCl to g AgCl, and that will be your theoretical yield. Your actual yield is 17.8 g, and so if you divide actual yield/theoretical yield * 100 you get your percent yield. You need molecular weights for this: MW AgNO3 = 170 g/mol MW AgCl = 143 g/mol theoretical yield: 21.8 g AgNO3 * (1 mol AgNO3 / 170 g AgNO3) * (1 mol AgCl / 1 mol AgNO3) * (143 g AuCl / mol AuCl) = 18.3 g Percent yield: 17.8… [cont.]
Answered by Chemis-T - Tue Jan 6 02:40:35 2009
Molarity questions! silver nitrate(I) react with 100mL of barium bromide solution...?
Q. An excess silver nitrate(I) react with 100mL of barium bromide solution to give 0.300g of solid compound. What is the net ionic equation for the reaction? Molarity of barium bromide solution? How many barium bromide ions present in original solutions? Kindly help me with this chemistry questions. Thanks
Asked by ABCDEFGHIJKLMNOPQRSTUVWXYZ - Sun Oct 18 03:28:39 2009 - - 1 Answers - 0 Comments
Q. An excess silver nitrate(I) react with 100mL of barium bromide solution to give 0.300g of solid compound. What is the net ionic equation for the reaction? Molarity of barium bromide solution? How many barium bromide ions present in original solutions? Kindly help me with this chemistry questions. Thanks
Asked by ABCDEFGHIJKLMNOPQRSTUVWXYZ - Sun Oct 18 03:28:39 2009 - - 1 Answers - 0 Comments
excess copper reacts with 8.32 g of silver nitrate. How much copper is needed? how much silver is produced?
Q. Please, help me with this question. And if you could explain it as well.
Asked by unknown - Mon Sep 14 04:07:25 2009 - - 1 Answers - 0 Comments
A. first of all get the equation: Cu + AgNO3-->Cu(No3)2 + Ag n(AgNO3) = m/M = 8.32/(107.9+13+3x16) = 8.32/169.9 = 0.04897 Since the silver nitrate is the limiting reagent, then that means 0.04897 moles of copper is needed since it is in a 1:1 ratio. Alos the amount of silver (Ag) produced is 0.04897moles since AgNO3 is the limiting reagent and AgNO3 and Ag are in a 1:1 ratio m(Ag)=nxM =0.04897 x 107.9 = 5.28 grams Or if you need the answer in mass than 5.28 grams of silver is produced :)
Answered by unknown - Mon Sep 14 04:25:59 2009
Q. Please, help me with this question. And if you could explain it as well.
Asked by unknown - Mon Sep 14 04:07:25 2009 - - 1 Answers - 0 Comments
A. first of all get the equation: Cu + AgNO3-->Cu(No3)2 + Ag n(AgNO3) = m/M = 8.32/(107.9+13+3x16) = 8.32/169.9 = 0.04897 Since the silver nitrate is the limiting reagent, then that means 0.04897 moles of copper is needed since it is in a 1:1 ratio. Alos the amount of silver (Ag) produced is 0.04897moles since AgNO3 is the limiting reagent and AgNO3 and Ag are in a 1:1 ratio m(Ag)=nxM =0.04897 x 107.9 = 5.28 grams Or if you need the answer in mass than 5.28 grams of silver is produced :)
Answered by unknown - Mon Sep 14 04:25:59 2009
What weight of silver chloride will be produced if 25cm3 of 0.1M silver nitrate is added to excess NaCl?
Q. What weight of silver chloride will be produced if 25cm3 of 0.1M silver nitrate is added to excess NaCl?
Asked by Joyce - Tue Oct 31 18:02:01 2006 - - 1 Answers - 0 Comments
A. AgNO3 + NaCl ---> AgCl Number of Moles of AgNO3: 0.0025 mol Molar Mass of AgCl: 143.321 g/mol Weight of AgCl = 0.358 grams
Answered by titanium007 - Tue Oct 31 21:12:26 2006
Q. What weight of silver chloride will be produced if 25cm3 of 0.1M silver nitrate is added to excess NaCl?
Asked by Joyce - Tue Oct 31 18:02:01 2006 - - 1 Answers - 0 Comments
A. AgNO3 + NaCl ---> AgCl Number of Moles of AgNO3: 0.0025 mol Molar Mass of AgCl: 143.321 g/mol Weight of AgCl = 0.358 grams
Answered by titanium007 - Tue Oct 31 21:12:26 2006
If 25.2 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.847 g of preci
Q. If 25.2 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.847 g of preci
Asked by Mary T - Fri Oct 6 19:10:20 2006 - - 2 Answers - 0 Comments
A. Your question got cut off, but I'm assuming the ppt. is silver chloride, and you want to know the concentration of silver nitrate? Convert your mass AgCl to moles Use the stoichiometry to get moles of AgNO3 (1:1 in this case) Then divide by the number of liters you started with. Or, if you want the mass of silver nitrate, convert back to grams. That's the procedure.
Answered by sleeptablets - Fri Oct 6 19:16:09 2006
Q. If 25.2 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.847 g of preci
Asked by Mary T - Fri Oct 6 19:10:20 2006 - - 2 Answers - 0 Comments
A. Your question got cut off, but I'm assuming the ppt. is silver chloride, and you want to know the concentration of silver nitrate? Convert your mass AgCl to moles Use the stoichiometry to get moles of AgNO3 (1:1 in this case) Then divide by the number of liters you started with. Or, if you want the mass of silver nitrate, convert back to grams. That's the procedure.
Answered by sleeptablets - Fri Oct 6 19:16:09 2006
how many moles of silver are produced?
Q. if 350g of Cu are reacted with excess silver nitrate according to the equation Cu(s) + 2AgNO3(aq) ---> 2Ag(s) + Cu(NO3)2 ?
Asked by Travvy - Sun Jun 1 22:44:35 2008 - - 2 Answers - 0 Comments
A. First, find the moles of Cu used. 350 g Cu/63.5 g Cu = 5.51 mol Cu Every mole of copper produces 2 moles of silver. 5.51 mol Cu*2 mol Ag = 11.02 mol Ag
Answered by The Duchess - Sun Jun 1 22:48:47 2008
Q. if 350g of Cu are reacted with excess silver nitrate according to the equation Cu(s) + 2AgNO3(aq) ---> 2Ag(s) + Cu(NO3)2 ?
Asked by Travvy - Sun Jun 1 22:44:35 2008 - - 2 Answers - 0 Comments
A. First, find the moles of Cu used. 350 g Cu/63.5 g Cu = 5.51 mol Cu Every mole of copper produces 2 moles of silver. 5.51 mol Cu*2 mol Ag = 11.02 mol Ag
Answered by The Duchess - Sun Jun 1 22:48:47 2008
If 27.0 mL of silver nitrat solution reacts w excess potassium chloride solution to yield 0.892 g of ?
Q. If 27.0 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.892 g of precipitate, what is the molarity of silver ion in the original solution? M
Asked by dora - Wed Oct 1 01:00:58 2008 - - 1 Answers - 0 Comments
A. Moles AgCl = 0.892 g / 143.32 = 0.00622 = moles Ag+ [Ag+] = 0.00622 / 0.027 L =0.231 M
Answered by Dr.A - Wed Oct 1 10:06:55 2008
Q. If 27.0 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.892 g of precipitate, what is the molarity of silver ion in the original solution? M
Asked by dora - Wed Oct 1 01:00:58 2008 - - 1 Answers - 0 Comments
A. Moles AgCl = 0.892 g / 143.32 = 0.00622 = moles Ag+ [Ag+] = 0.00622 / 0.027 L =0.231 M
Answered by Dr.A - Wed Oct 1 10:06:55 2008
What is the percent yield for this reaction?
Q. A 5.00 g piece of copper is placed in a solution of silver nitrate (AgNO3) in which there is excess silver nitrate. The product produced in this reaction is copper nitrate, Cu(NO3)2. The mass of the silver produced is 15.2g. What is the percent yield for this reaction?
Asked by Maddie L - Sun Jun 3 17:07:00 2007 - - 1 Answers - 0 Comments
A. You are asking for the percent yield of silver, right? Here's the equation to find the theoretical yield: 5.00g((1 mol Cu)/(63.546g))((2 mol Ag)/(1 mol Cu))((107.868g)/(1 mol Cu(NO3)2)) = 16.975 g Percent yield = actual/theoretical= 15.2/16.975 = 89.5%
Answered by buccaneerjoe - Sun Jun 3 17:21:50 2007
Q. A 5.00 g piece of copper is placed in a solution of silver nitrate (AgNO3) in which there is excess silver nitrate. The product produced in this reaction is copper nitrate, Cu(NO3)2. The mass of the silver produced is 15.2g. What is the percent yield for this reaction?
Asked by Maddie L - Sun Jun 3 17:07:00 2007 - - 1 Answers - 0 Comments
A. You are asking for the percent yield of silver, right? Here's the equation to find the theoretical yield: 5.00g((1 mol Cu)/(63.546g))((2 mol Ag)/(1 mol Cu))((107.868g)/(1 mol Cu(NO3)2)) = 16.975 g Percent yield = actual/theoretical= 15.2/16.975 = 89.5%
Answered by buccaneerjoe - Sun Jun 3 17:21:50 2007
What are the half reactions for Silver Nitrate(aq) and Iron(s)?
Q. Also, how would I make sure that the Iron used would be in excess? Thanks!
Asked by Axiz - Tue Feb 9 21:08:50 2010 - - 1 Answers - 0 Comments
A. Ag+ + 1e- = Ag Fe = Fe2+ + 2e-
Answered by Dr.A - Wed Feb 10 09:46:13 2010
Q. Also, how would I make sure that the Iron used would be in excess? Thanks!
Asked by Axiz - Tue Feb 9 21:08:50 2010 - - 1 Answers - 0 Comments
A. Ag+ + 1e- = Ag Fe = Fe2+ + 2e-
Answered by Dr.A - Wed Feb 10 09:46:13 2010
What is the mass percent of sodium chloride?
Q. A 0.8000 g sample of impure sodium choride is dissolved in water, and the chloride is precipitated with excess silver nitrate, producing 1.000 g of silver chloride. What is the mass percent of sodium chloride in the original impure sample? a. ) 40.77 b. ) 24.74 c. ) 50.97 d. ) 80.00 e. ) 0.6977 I worked it out and I got e). Is that correct?
Asked by secondaryhoper - Sat Sep 22 19:29:39 2007 - - 1 Answers - 0 Comments
A. Atomic weights: Ag=108 Cl=35.5 Na=23 AgCl=143.5 NaCl=58.5 NaCl + AgNO3 ===> AgCl + NaNO3 1.000gAgCl x 1molAgCl/143.5gAgCl x 1molNaCl/1molAgCl x 58.5gNaCl/1molNaCl = 0.4077g NaCl 0.4077/0.8000 x 100% = 50.96% Looks like c.)
Answered by steve_geo1 - Sat Sep 22 19:48:34 2007
Q. A 0.8000 g sample of impure sodium choride is dissolved in water, and the chloride is precipitated with excess silver nitrate, producing 1.000 g of silver chloride. What is the mass percent of sodium chloride in the original impure sample? a. ) 40.77 b. ) 24.74 c. ) 50.97 d. ) 80.00 e. ) 0.6977 I worked it out and I got e). Is that correct?
Asked by secondaryhoper - Sat Sep 22 19:29:39 2007 - - 1 Answers - 0 Comments
A. Atomic weights: Ag=108 Cl=35.5 Na=23 AgCl=143.5 NaCl=58.5 NaCl + AgNO3 ===> AgCl + NaNO3 1.000gAgCl x 1molAgCl/143.5gAgCl x 1molNaCl/1molAgCl x 58.5gNaCl/1molNaCl = 0.4077g NaCl 0.4077/0.8000 x 100% = 50.96% Looks like c.)
Answered by steve_geo1 - Sat Sep 22 19:48:34 2007
Calculate the percent mass of each mixture component?
Q. A sample of a mixture containing only sodium chloride and potassium chloride has a mass of 4.000g. When this sample is dissolved in water and excess silver nitrate is added, a white solid (Sliver Chloride) forms. After filtration and drying, the solid silver chloride has a mass of 8.5904g. Calculate the mass percent of each mixture component. Any help would be much appreciated, I am working on my AP Chem summer work and have been stuck on this problem for over an hour, thanks!
Asked by Charlie A - Wed Aug 12 12:20:50 2009 - - 2 Answers - 0 Comments
A. moles AgCl = 8.5904 g/ 143.32 g/mol= 0.05994 = moles Cl- in the mixture let x = mass NaCl let y = mass KCl x + y = 4.000 ==> x = 4.000 - y x / 58.44 + y / 74.55 = 0.05994 4.000 -y/ 58.44 + y / 74.55 = 0.05994 298.2 - 74.55y +58.44 y = 261.1 37.1 = 16.11 y y = 2.303 g = mass KCl x = 4.000 - 2.303 =1.697 g = mass NaCl % KCl = 2.303 x 100/ 4.000 =57.58 % NaCl =42.42
Answered by Dr.A - Wed Aug 12 12:31:20 2009
Q. A sample of a mixture containing only sodium chloride and potassium chloride has a mass of 4.000g. When this sample is dissolved in water and excess silver nitrate is added, a white solid (Sliver Chloride) forms. After filtration and drying, the solid silver chloride has a mass of 8.5904g. Calculate the mass percent of each mixture component. Any help would be much appreciated, I am working on my AP Chem summer work and have been stuck on this problem for over an hour, thanks!
Asked by Charlie A - Wed Aug 12 12:20:50 2009 - - 2 Answers - 0 Comments
A. moles AgCl = 8.5904 g/ 143.32 g/mol= 0.05994 = moles Cl- in the mixture let x = mass NaCl let y = mass KCl x + y = 4.000 ==> x = 4.000 - y x / 58.44 + y / 74.55 = 0.05994 4.000 -y/ 58.44 + y / 74.55 = 0.05994 298.2 - 74.55y +58.44 y = 261.1 37.1 = 16.11 y y = 2.303 g = mass KCl x = 4.000 - 2.303 =1.697 g = mass NaCl % KCl = 2.303 x 100/ 4.000 =57.58 % NaCl =42.42
Answered by Dr.A - Wed Aug 12 12:31:20 2009
From Yahoo Answer Search: 'excess silver nitrate'
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