How many grams of water are generated when 5.6 liters of propane are burned?

To solve, we write down the equation of the process:

С3Н8 + 5О2 = 3СО2 + 4Н2О + Q – propane combustion, carbon dioxide and water are released;
Let’s make the calculations:
M (C3H8) = 44 g / mol;

M (H2O) = 18 g / mol.

3. Proportion:

1 mol of gas at normal level – 22.4 liters;

X mol (C3H8) – 5.6 liters from here, X mol (C3H8) = 1 * 5.6 / 22.4 = 0.25 mol;

0.25 mol (C3H8) – X mol (H2O);

-1 mol – 4 mol from here, X mol (H2O) = 0.25 * 4/1 = 1 mol.

Find the mass of the product:
m (H2O) = Y * M = 1 * 18 = 18 g

Answer: the mass of water is 18 g