They burned 7 liters of methane in 8 liters of oxygen. Find the volume and mass of the resulting gas, and the mass of water.

Let’s implement the solution:
1. According to the condition of the problem, we compose the equation:
CH4 + 2O2 = CO2 + 2H2O + Q – the combustion reaction is accompanied by the release of heat, carbon dioxide and water;
2. Calculations:
M (CH4) = 16 g / mol;
M (CO2) = 44 g / mol;
M (H2O) = 18 g / mol.
3. Proportions:
1 mole of gas at normal level – 22.4 liters;
X mol (CH4) – 7 liters. hence, X mol (CH4) = 1 * 7 / 22.4 = 0.31 mol (the substance is in short supply);
Y (CO2) = 0.31 mol since the quantities of these substances are equal to 1 mol according to the equation.
1 mole of gas at normal level – 22.4 liters;
X mol (O2) – 8 liters. hence, X mol (O2) = 1 * 8 / 22.4 = 0.36 mol;
The calculation is made for the substance in deficiency.
0.31 mol (CH4) – X mol (H2O);
-1 mol – 2 mol from here, X mol (H2O) = 0.31 * 2/1 = 0.62 mol.
4. Find the masses of products:
m (H2O) = Y * M = 0.62 * 18 = 11.16 g;
m (CO2) = Y * M = 0.31 * 44 = 13.64 g.
Answer: in the course of the reaction formed: water weighing 11.16 g and carbon monoxide – 13.64 g.