How many grams of water will be obtained by burning 11.2 liters of methane?

Let’s implement the solution:
1. Let’s compose the reaction equation:
СН4 + 2О2 = СО2 + 2Н2О – methane combustion reaction, carbon dioxide and water are released;
2. Determine the molecular weights of substances:
M (CH4) = 12 + 4 = 16 g / mol;
M (H2O) = 1 * 2 + 16 = 18 g / mol;
3. Let’s calculate the number of moles of methane, if its volume is known:
1 mol of CH4 gas at n. y – 22.4 l;
X mol – 11.2 liters. hence, X mol (CH4) = 1 * 11.2 / 22.4 = 0.5 mol;
4. Let’s make the proportion:
0.5 mol (CH4) – X mol (H2O);
-1 mol – 2 mol from here, X mol (H2O) = 0.5 * 2/1 = 1 mol;
5. Let’s calculate the mass of water:
M (H2O) = Y * M = 1 * 18 = 18 g.
Answer: 18 g of H2O is released during the reaction.