$10\text{g}$ of hydrogen and $64\text{g}$ of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:

Reaction: $2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$ Moles of ${\text{H}}_2=10/2=5\text{mol}$.Moles of ${\text{O}}_2=64/32=2\text{mol}$.Stoichiometry requires $2$ moles of ${\text{H}}_2$ for $1$ mole of ${\text{O}}_2$.So $2$ moles of ${\text{O}}_2$ require $4$ moles of ${\text{H}}_2$.Since we have $5$ moles of ${\text{H}}_2$, Oxygen is the limiting reagent.Moles of ${\text{H}}_2\text{O}$ formed $=2\times {\text{moles of O}}_2=2\times 2=4\text{mol}$.