10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be

2H₂​+O_2​→2H_2​O

10 g of hydrogen (molar mass 2 g/mol) = $\frac{10 \mathrm{g}}{{\displaystyle \raisebox{1ex}{$2 \mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}$ ​= 5 mol 

 64 of oxygen  (molar mass 32 g/mol) = $\frac{64 \mathrm{g}}{{\displaystyle \raisebox{1ex}{$32 \mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}$ ​= 2 mol 

As per the balanced chemical equation, 1 mole of oxygen will react with 2 moles of hydrogen to form 2 moles of water.

2 moles of oxygen will react with 2 × 2 = 4 moles of hydrogen to form 4 moles of water.

Here, oxygen is the limiting reagent and hydrogen is the excess reagent.