Question

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

Answer 1

Answer:

                  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