When 4g of hydrogen is reacts with 4g of Oxygen gas, the amount of water produced in the

reaction is ${\text{2H}}_{2\left(\text{g}\right)}+{\text{O}}_{2\left(\text{g}\right)}\to 2{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$

$2{\text{H}}_2+{\text{ O}}_{\text{2}}\to 2{\text{H}}_{\text{2}}\text{O}$

$\text{2×2=4g 32g 2×18=36g}$

${\text{4g of H}}_2{\text{ required 32g of O}}_2$

Here to react with 4g of Hydrogen only 4g of oxygen is only available

Hence the limiting reagent is Oxygen.

$\underset{(\text{excess)}}{4\text{g}}\underset{\left(\text{limiting}\right)}{\text{4g}}\text{ ?}$

${\text{32g of O}}_2\to 36g{\text{\hspace{0.17em}of H}}_{\text{2}}O$

${\text{4g of O}}_2\to ?\text{X }g{\text{\hspace{0.17em}of H}}_{\text{2}}O$

$\text{X=}\frac{\text{4×36}}{32}=4.5g$