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Offline oceanmd

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Please check these problems
« on: February 27, 2008, 12:46:16 AM »
Thank you very much.
1. For reaction 2Na+Cl2=2NaCl, how many grams of NaCl could be produced from 103.0 g of Na and 13.0 L of Cl2?
My answer:
103 g Na=4.48mol Na
13L Cl=0.58mol Cl2
Cl2 is the limiting reagent
0.58mol Cl2=1.16mol NaCl=67.79g NaCl

2. In a particular reaction between copper metal and silver nitrate, 12.7g Cu produces 38.1g Ag. What is the percent yield of silver in this reaction? Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
My answer: 12.7 g Cu = 0.19998 mol Cu = 0.3997 mol Ag = 43.116 g Ag (theoretical yield), 38.1/43.116 x 100% = 88.37%

3. Hydrogen gas is produced when Zinc reacts with hydrochloric acid. How many grams of zinc are needed to produce 112L of H2 at STP? Zn + 2HCl = ZnCl2 + H2
My answer: 112 L H2 = 5 mol H2 = 5 mol Zn = 326.95 g Zn

4. Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 9.9 g Pb(NO3)2 is heated to give 5.5 g of PbO? 2Pb(NO3)2 = 2PbO + 4NO2 + O2 
My answer: 82.41% yield of PbO
 
5. Metallic copper is formed when aluminum reacts with copper (II) sulfate. How many grams of metallic copper can be obtained when 54.0 g of Al reacts with 319g of CuSO4? 2Al + 3CuSO4 = Al2(SO4)3 + 3Cu
My answer: CuSO4-limiting reagent, 127.09 g Cu

6. How many moles of aluminum are needed to react completely with 1.2 mol of FeO? 2Al + 3FeO = 3Fe + Al2O3
My answer: 0.8 mol Al

7. How many moles of glucose C6H12O6 can be burned biologically when 10.0 mol of oxygen is available? C6H12O6 + 6O2 = 6CO2 + 6H2O
My answer: 1.667 mol C6H12O6

8. How many liters of NH3, at STP, will react with 5.3 x 10^34 molecules of O2 to form NO2 and water? 4NH3 + 7O2 = 4NO2 + 6H2O
My answer: 1.13 x 10^12 L NH3

9. Iron (III) oxide is formed when iron combines with oxygen in the air. How many grams of Fe2O3 are formed when 16.7 g of Fe reacts completely with oxygen? 4Fe + 3O2 = 2Fe2O3
My answer: 23.95 g Fe2O3

10. Hydrogen gas can be produced by reacting aluminum with sulfuric acid. How many molecules of sulfuric acid are needed to completely react with 15.0 mol of aluminum? 2Al + 3H2SO4 = Al2(SO4)3 + 3H2
My answer: 1.35 x 10^25 molecules H2SO4

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