HONORS CHEMISTRY STOICHIOMETRY PROBLEMS WORKSHEET 1 NAME: DATE: S. The unbalance
ID: 1074225 • Letter: H
Question
HONORS CHEMISTRY STOICHIOMETRY PROBLEMS WORKSHEET 1 NAME: DATE: S. The unbalanced decomposition reaction of butane gas in excess oxygen produces carbon dioxide gas and water vapor. C.Hon + Oe===> Co, u + HOHo. Starting with 11.6 grams of butane, how many grams of carbon dioxide gas and water vapor are formed? 6. The burning of solid sulfur in air produces sulfur dioxide gas. Balance the reaction. How many moles and molecules of sulfur dioxide does the burning of 3 moles of sulfur form 7. The Haber reaction produces ammonia, an important nitrogenous compound needed to make plant fertilizers. The unbalanced reaction is: N, + H210--> NH, + heat. Is the reaction exothermic or endothermic? If 170.0 grams of ammonia are produced, then how many grams of nitrogen gas and hydrogen gas are needed? How many molecules of each reactant are needed?
Explanation / Answer
5.
2C4H10(g) + 13O2(g) -------------> 8Co2(g) + 10H2O
2 moles of butane on combustion to gives 8 moles of Co2
2*58g of butane on combustion to gives 8*44g of CO2
11.6g of butane on combustion to gives = 8*44*11.6/2*58 = 35.2g of Co2
2 moles of butane on combustion to gives 10 moles of H2O
2*58g of butane on combustion to gives 10*18g of H2O
11.6g of butane on combustion to gives = 10*18*11.6/2*58 = 18.g of H2O
6. S + O2 ---------> SO2
1 mole of S burn to gives 1 mole of So2
3 moles of S burn to gives 3 moles of So2
no of molecules = no of moles* 6.023*10^23
= 3*6.023*10^23
= 1.8069*10^24 molecules
7. N2(g) + 3H2(g) ----------> 2NH3(g) + heat
It is exothermic reaction
2 moles of NH3 produced from 1 mole of N2
2*17g of NH3 produced from 28g of N2
170g of NH3 produced from = 28*170/2*17 = 140g of N2
2 mole of NH3 produced from 3 moles of H2
2*17g of NH3 produced from 3*2g of H2
170g of NH3 produced from = 3*2*170/2*17 = 30g of H2
no of molecules of N2 = no of moles * 6.023*10^23
= 140*6.023*10^23/28 = 3.0115*10^24 of N2 molecules
no of moleculesof H2 = no of moles * 6.023*10^23
= 30*6.023*10^23/2 = 9.0345*10^24 of H2 molecules
8. 2Na(s) + 2H2O(l) -----------> 2NaOH(aq) + H2(g)+ light
1 mole of H2 produced from 2 moles of Na
2g of H2 produced from 2*23g of Na
44.8g of H2 produced from = 2*23*44.8/2 = 1030.4g of Na
1 mole of H2 produced from 2 moles of H2O
2g of H2 produced from 2*18g of H2O
44.8g of H2 produced from = 2*18*44.8/2 = 806.4g of H2O
9. 2H2(g) + O2(g) -----------> 2H2O
2 moles of H2O produced from 2 moles of H2
2*18g of H2O produced from 2moles of H2
11.2g of H2O produced from = 2*11.2/2*18 = 0.62 moles of H2
2 moles of H2O produced from 1 moles of O2
2*18g of H2O produced from 1 mole of O2
11.2g of H2O produced from = 1*11.2/2*18 = 0.31 moles of O2
no of molecules of H2 = no of moles* 6.023*10^23
= 0.62*6.023*10^23 = 3.73*10^23 molecules of H2
no of molecules of O2 = no of moles* 6.023*10^23
= 0.31*6.023*10^23
= 1.86*10^23 molecules of O2
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