Determine the volume of СО2 that will be released during combustion of 4.48 liters of С2Н6.

1. Let’s compose the reaction equation for ethane combustion:
2C2H6 + 7O2 = 4CO2 + 6H2O;
2. Let’s find the amount of substance in ethane:
4.48 / 22.4 = 0.2 mol;
3. Before carbon dioxide there is a 2 times higher coefficient than before ethane. This means that the amount of substance before carbon dioxide is 0.4 mol. Let’s find the volume of carbon dioxide:
22.4 * 0.4 = 8.96 liters;
Answer: The volume of carbon dioxide is 8.96 liters.