What volume of carbon dioxide will be released when burning benzene weighing 1 g, containing 0.2% of impurities

What volume of carbon dioxide will be released when burning benzene weighing 1 g, containing 0.2% of impurities, if the yield of the reaction product is 65% of the theoretically possible.

Let’s find the mass of pure benzene, without impurities.

100% – 0.2% = 99.8%.

1g – 100%,

m – 99.8%.

m = (1 g × 99.8%): 100% = 0.998 g.

Let’s find the amount of benzene substance C6H6.

n = m: M.

M (C6H6) = 78 g / mol.

n = 0.998 g: 78 g / mol = 0.01 mol.

Let’s find the quantitative ratios of substances.

2С6Н6 + 15О2 → 12СО2 + 6Н2О

For 2 mol of С6Н6, there are 12 mol of СO2.

Substances are in quantitative ratios 2: 12 = 1: 6.

The amount of CO2 will be 6 times more than C6H6.

n (CO2) = 6n (C6H6) = 0.01 × 6 = 0.06 mol.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.06 mol × 22.4 L / mol = 1.344 L.

1.344 received by theory, by calculations. According to the condition of the problem, the CO2 yield was 65% of the theoretically possible.

1,344 l – 100%,

V – 65%.

V = (1.344 L × 65%): 100% = 0.87 L.

Answer: 0.87 liters.