Chemistry - Redox Reactions Question with Solution | TestHub

${\mathrm{H}}_2+1/2{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}$ $2\mathrm{g}16\mathrm{g}18\mathrm{g}$ $10\mathrm{g}{\mathrm{H}}_2$ required ${\mathrm{O}}_2=80$ which is not present 64 g ${\mathrm{O}}_2$ required $8\mathrm{g}$ of ${\mathrm{H}}_2$ and ${\mathrm{H}}_2$ left $=2\mathrm{g}$. Thus, ${\mathrm{O}}_2$ is the limiting reactant and ${\mathrm{H}}_2$ is excess reactant. Hence, ${\mathrm{H}}_2\mathrm{O}$ formed from 64 of ${\mathrm{O}}_2$ $$\begin{array}{rl}& =\frac{18}{16}\times 64\\ & =72\mathrm{g}=\frac{72}{18}\mathrm{mole}\\ & =4\mathrm{mole}\end{array}$$