Find the volume of carbon dioxide that is released during the combustion of the 20th acytene if the yield is 90%.

Combustion of acetylene: 2C2H2 + 5O2 = 4CO2 + 2H2O.
1) Molar mass of acetylene: 12 * 2 + 1 * 2 = 26.
2) The amount of acetylene substance: 20/26 = 0.8 mol.
3) According to the reaction equation, there are four moles of carbon dioxide for two moles of acetylene. This means that the amount of carbon dioxide substance is 2 * 0.8 / 4 = 0.4 mol.
4) The volume of carbon dioxide: 0.4 * 22.4 = 8.96 liters – theoretical output.
5) Finding a practical way out. To do this, multiply the volume of carbon dioxide by the percentage of practical output and divide by one hundred percent: 8.96 * 90/100 = 8.064 liters – the answer.