The composition of natural gases that are used as fuel can include ethane C2H6. According to the combustion

The composition of natural gases that are used as fuel can include ethane C2H6. According to the combustion equation C2H6 + 7O2 = 4C02 + 6H2O. Calculate: the amount of substance, mass and volume of oxygen, which was spent on the combustion of ethane with a volume of 3.2 liters.

For the solution, we write the data:
V = 3.2 liters. X g -? X l. -? Y -?
1. 2С2Н6 + 7О2 = 4СО2 + 6Н2О + Q – combustion is accompanied by the release of heat, carbon monoxide (4), water;
2. Determine the amount of the original substance:
1 mol of gas at normal level – 22.4 liters;
X mol (C2H6) – 3.2 liters. hence, X mol (C2H6) = 1 * 3.2 / 22.4 = 1.428 mol
3. Proportion:
X mol (O2) – 1.428 mol (C2H6);
-7 mol – 2 mol from here, X mol (O2) = 7 * 1.428 / 2 = 4.998 mol = 5 mol.
4. Find the mass, volume of O2:
m (O2) = Y * M = 5 * 32 = 160 g;
V (O2) = 5 * 22.4 = 112 liters.
Answer: for the combustion process, oxygen with a mass of 160 g and a volume of 112 liters is required.