What is the volume of carbon dioxide formed during the combustion of ethylene, weight 80 grams

What is the volume of carbon dioxide formed during the combustion of ethylene, weight 80 grams if it contains 10% impurities?

1. Let’s write down the reaction equation:

C2H4 + 3O2 = 2CO2 + 2H2O.

2. Find the amount of ethylene:

ω (C2H4) = 100% – ω (impurities) = 100% – 10% = 90%.

m (C2H4) = (m (mixture) * ω (C2H4)) / 100% = 80 g * 90% / 100% = 72 g.

n (C2H4) = m (C2H4) / M (C2H4) = 72 g / 28 g / mol = 2.57 mol.

3. According to the reaction equation, we find the amount, and then the volume of carbon dioxide (Vm – molar volume, constant equal to 22.4 l / mol):

n (CO2) = 0.5n (C2H4) = 0.5 * 2.57 mol = 1.29 mol.

V (CO2) = n (CO2) * Vm = 1.29 mol * 22.4 L / mol = 28.8 L.

Answer: V (CO2) = 28.8 liters.