$2{\text{H}}_{2 \left(g\right)}+{\text{O}}_{2 \left(g\right)}\to 2{\text{H}}_2{\text{O}}_{\left(l\right)}$ . Based on this reaction, how much of oxygen is required to combine 11.2 L of Hydrogen?

2 moles of Hydrogen means 44.8 L of hydrogen gas, in the above equation 44.8 L

of hydrogen combines with 32g of Oxygen gas to from 2 moles of water.

Hence 11.2 L of hydrogen can combine with

$\frac{11.2}{44.8}\times 32$  

$=8g$ .