How many liters of water can be produced from 5.0 liters of butane gas at STP, assuming excess oxygen?

25 L of H₂O

Solution:

The balanced chemical equation is as,

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

As at STP one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,

44.8 L (2 mol) C₄H₁₀ produces = 224 L (10 mol) of H₂O

So,

5.0 L of C₄H₁₀ will produce = X L of H₂O

Solving for X,

X = (5.0 L * 224 L) : 44.8 L

X = 25 L of H₂O