Burned 100 liters of methane calculate the mass of oxygen and carbon dioxide in the reaction.

Let’s implement the solution:
1. We compose the equation according to the condition of the problem:
V = 100 HP X g -? X g -?
СН4 + 2О2 = СО2 + 2Н2О + Q – combustion is accompanied by the release of heat, carbon dioxide, water;
2. We make calculations:
M (O2) = 32 g / mol;
M (CO2) = 44 g / mol.
3. Determine the amount of the original substance:
1 mol of gas at normal level – 22.4 liters;
X mol (CH4) – 100 liters. hence, X mol (CH4) = 1 * 100 / 22.4 = 4.46 mol.
4. Proportion:
4.46 mol (CH4) – X mol (CO2);
-1 mol                  -2 mol hence, X mol (CO2) = 4.46 * 2/1 = 8.92 mol.
5. Find the mass of the product:
m (CO2) = Y * M = 8.92 * 32 = 285.4 g.
Answer: carbon dioxide was released with a volume of 285.4 g.