Question

2 C₂H_6(g)      +      2 O_2(g) =  CO_2(g)    +      2 H₂O_(I)

1. How many grams of H₂O will be produced by 58.2 L of CH₄ at STP? Assume an excess of O₂.

Answer 1

We use the fact that 1 mole of any gas at STP has a volume of 22.4 L to find how much water was produced:

$\text{58.2 L of CH}_4 \text{ at STP} \times\cfrac{\text{1 mol of CH}_4}{\text{22.4 L of CH}_4}\times\cfrac{\text{2 mol of H}_2O}{\text{1 mol of CH}_4}\times\cfrac{\text{18 g of H}_2O}{\text{1 mol of H}_2O} \\ =93.54\text{ g of H}_2O$

In conclusion, 93.54 g of water was produced during the combustion.

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.