12. Ozone, O, is a produced in automobile exhaust by the reaction represented by
ID: 559558 • Letter: 1
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12. Ozone, O, is a produced in automobile exhaust by the reaction represented by the equation Nola) + 02(g) OO(g) + Orlg). What mass of ozone is predicted to form from the reaction of 10.0 g NO2 in a car's exhaust and excess oxygen? A. 5.5g0 B. 9.08O C. 10.5gO 13. The combustion of 100 g of butane (CaHso) produces 280 g of carbon dioxide. What is the percent yield, if the theoretical yield is 303 g of CO? A.9.23 % B. 92.3% C. 108 % 14. What is the percent yield of a reaction if the product recovered is 21.2 g and the theoretical yield is 28.4g? A. 74.6% B. 134% C. 0.746% . 15, what is the actual yield of a reaction if the percent yield is 80.6 % of the 42.4 g theoretical yield? A. 52.68 B. 42.4g C. 342g 16. Iron metal reacts with oxygen in the air to form rust, according to the balanced equation shown below. How many moles of oxygen are needed to react with 0.100 mole of Fe? 4 Fe (s)+3 022 Fe Os (s) A. 0.300 mole B. 0.133 mole C. 0.075 mole 17. Iron metal reacts with oxygen in the air to form rust, according to the balanced equation shown below. How many grams of oxygen are needed to react with 5.85 g of Fe? 4 Fe (s) + 3 0282 Fe0; (s)Explanation / Answer
12)
Molar mass of NO2 = 1*MM(N) + 2*MM(O)
= 1*14.01 + 2*16.0
= 46.01 g/mol
mass of NO2 = 10 g
mol of NO2 = (mass)/(molar mass)
= 10/46.01
= 0.2173 mol
From balanced chemical reaction, we see that
when 1 mol of NO2 reacts, 1 mol of O3 is formed
mol of O3 formed = moles of NO2
= 0.2173 mol
Molar mass of O3 = 48 g/mol
mass of O3 = number of mol * molar mass
= 0.2173*48
= 10.5 g
Answer: C
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