Question

What volume of O₂ would be required to burn 224 mL of H₂ at 273K and 1.013 bar?

2H₂ + O₂ ^→ 2H₂O

Answer 1

273 K and 1.013 bar are STP conditions, so 1 mol of gas occupies volume of 22.4 L

n(H₂) $= \frac{0.224}{22.4} = 0.01 \;mol$

According to the reaction:

n(O₂) $= \frac{1}{2}n(H_2) = \frac{1}{2} \times 0.01 = 0.005 \;mol$

V(O₂) $= 0.005 \times 22.4 = 0.112 \;L = 112 \;mL$

Answer: 112 mL