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$56 \mathrm{~g}$ of nitrogen and $8 \mathrm{~g}$ of hydrogen gas are heated in a closed vessel. At equilibrium $34 \mathrm{~g}$ of ammonia are present. The equilibrium number of moles of nitrogen, hydrogen and ammonia are respectively
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$1,1,2$
$\therefore$ Mass of $N_{2}$ left $=56-28=28$ g
$\therefore$ Moles of $\mathrm{N}_{2}$ left $=\frac{28}{28}=1$ mole
Mass of $\mathrm{H}_{2}$ left $=8-6=2 \mathrm{~g}$
$\therefore$ Moles of $\mathrm{H}_{2}$ left $=\frac{2}{2}=1$ mole
Moles of ammonia $=\frac{34}{17}=2$ moles
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