What is the mass of water produced by the combustion of 5 liters of butane?

To solve the problem, we compose the reaction equation:
2С4Н10 + 13О2 = 8СО2 + 10Н2О – butane combustion reaction, carbon dioxide and water are released;
Let’s calculate the molecular weights of substances:
M (C4H10) = 4 * 12 + 1 * 10 = 58 g / mol;
M (H2O) = 18 g / mol;
Determine the amount of moles of butane:
1 mol of gas at n. y – 22.4 l;
X mol (C4H10) – 5 liters. hence, X mol (C4H10) = 1 * 5 / 22.4 = 0.22 l;
Let’s make the proportion:
0.22 mol (C4H10) – X mol (H2O);
-2 mol -10 mol from here, X mol (H2O) = 0.22 * 10/2 = 1.1 mol;
Find the mass of water: m (H2O) = Y * M = 1.1 * 18 = 19.8 g;
Answer: during the reaction, water with a mass of 19.8 g is released.