# In the synthesis of ammonia from nitrogen and hydrogen gases if 6×10^-2 mole of hydrogen disappears in 10 min the no. of moles of ammonia formed in 0.3 min is

The reaction can be written as :

${N}_{2}+3{H}_{2}\leftrightarrow 2N{H}_{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Rateofdissapearanceofhydrogen=Rateofappearanceofammonia\phantom{\rule{0ex}{0ex}}Thus,\phantom{\rule{0ex}{0ex}}\frac{1}{3}\times \frac{6\times {10}^{-2}}{10}=\frac{1}{2}\times \frac{{n}_{N{H}_{3}}}{0.3}\phantom{\rule{0ex}{0ex}}So,\phantom{\rule{0ex}{0ex}}{n}_{N{H}_{3}}=\frac{0.3}{3}\times \frac{2\times 6\times {10}^{-2}}{10}=1.2\times {10}^{-3}$

Thus, 1.2 mmol of ammonia are formed in 0.3 min.

Hope it helps.

Regards

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