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

$2{\mathrm{H}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}.$

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.