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If $15.0\ \mathrm{mol}$ of nitrogen are reacted with $30.0\ \mathrm{mol}$ of hydrogen, how much ammonia will be produced?

$$\ce{N2 + 3H2 -> 2NH3}$$

What I’ve tried is the following:

$$\dfrac{n(\ce{N2})}{n(\ce{H2})}=\dfrac{1}{3}=\dfrac{15.0\ \mathrm{mol}}{x}$$

$$\dfrac{n(\ce{N2})}{n(\ce{H2})}=\dfrac{1}{3}=\dfrac{x}{30\ \mathrm{mol}}$$

I’m not sure how to proceed from here. Any help would be appreciated.

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In the reaction:

$$\ce{N_2 + 3H_2 \to 2NH_3}$$

If 15 moles of $\ce{N_2}$ is used along with 30 moles of $\ce{H_2}$, then $\ce{H_2}$ will be the limiting reagent (Reactant which is consumed completely). This is because 15 moles of $\ce{N_2}$ will need 45 moles of $\ce{H_2}$ ($15 \times 3$, by the balanced reaction) for reacting completely, but as there is only 30 moles available hence it will react in a fixed proportion, hence you will have to assume the reacting moles of $\ce{N_2}$ as $x$. By this the equation goes,

$$\frac{\ce{N_2}}{\ce{H_2}}=\frac{1}{3}=\frac{x}{30}$$ From this $$x=\frac{30}{3}=10$$

Hence from 15 moles of $\ce{N_2}$ only 10 moles will react, hence the amount of $\ce{NH_3}$ formed will be 20 moles ($10 \times 2$, by the balanced reaction).

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