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Question

Find out the weight of ammonia formed when $$10$$ moles of $$N_2$$ reacts with $$20$$ moles of $$H_2$$.


Solution

     $$N_2+3H_2\longrightarrow 2NH_3$$                     $$N_2:H_2:NH_3$$
       $$1$$         $$3$$                 $$2$$ $$\rightarrow$$ Mole ratio is $$1:$$   $$3:$$    $$2$$
      $$10$$       $$30$$
Here $$H_2$$ is limiting as $$10$$ moles of $$N_2$$ can react with $$30$$ moles of $$H_2$$ .
Hence only $$\cfrac {20}{3}=6.67$$ moles of $$N_2$$ reacts to produce $$2\times 6.67$$ moles of $$NH_3$$
$$=13.44$$ moles of $$NH_3$$
$$=13.44 \times 17$$
$$=228.48g$$ of $$NH_3$$ is produced.

Chemistry

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