10 ml ${\mathrm{N}}_2$ is reacted with 20 ml ${\mathrm{H}}_2$ to form ${\mathrm{NH}}_3$.The correct statements is/are

${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$

Above reaction states that 1 vol of N₂ reacts with 3 vol of H₂ to give 2 vol of NH₃.

But given volume of ${\mathrm{H}}_2=20 \mathrm{mL};$ this means ${\mathrm{H}}_2$ is a limiting reagent.

Volume of NH₃ formed from 20 mL of ${\mathrm{H}}_2$.

$20\mathrm{mL} {\mathrm{H}}_2=\frac{2}{3}\times 20=13.3 \mathrm{mL} {\mathrm{NH}}_3$

Volume of ${\mathrm{N}}_2$ required for 20 mL of ${\mathrm{H}}_2$.

$20\mathrm{mL} {\mathrm{H}}_2=\frac{1}{3}\times 20=6.6 \mathrm{mL} {\mathrm{N}}_2 \mathrm{reacted}$

Volume of N₂ unreacted:

$\mathrm{Vol} \left(\mathrm{unreacted}\right)=10 \mathrm{mL}-6.6 \mathrm{mL}=3.4 \mathrm{mL}$