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Question

The balanced chemical equation for the formation of ammonia gas by the reaction between nitrogen gas an hydrogen gas is given.
$$N_2 + 3H \rightarrow 2NH_3$$
How many moles of ammonia are formed when $$6$$ moles of $$N_2$$ react with $$6$$ moles of $$H_2$$?


Solution

Given reaction for the formation of ammonia-
$${N}_{2} + 3 {H}_{2} \longrightarrow 2 N{H}_{3}$$
From the above reacton,
$$1$$ mole of $${N}_{2}$$ reacts with $$3$$ moles of $${H}_{2}$$.
Therefore,
$$6$$ moles of $${N}_{2}$$ will react with $$18$$ moles of $${H}_{2}$$.
Given amount of $${H}_{2} = 6 \text{ moles}$$
Therefore, $${H}_{2}$$ is the limiting reagent.
Therefore,
Again from the above reaction.
No. of moles of ammonia formed when $$3$$ moles of $${H}_{2}$$ react with excess of $${N}_{2} = 2$$
$$\therefore$$ No. of moles of ammonia formed when $$6$$ moles of $${H}_{2}$$ react with excess of $${N}_{2} = \cfrac{2}{3} \times 6 = 4 \text{ moles}$$
Hence $$4$$ moles of ammonia are formed when $$6$$ moles of $${N}_{2}$$ react with $$6$$ moles of $${H}_{2}$$.

Chemistry

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