Question

Ammonia is produced synthetically by the reaction below. How many moles of NH3 are formed when 200.0g of N2 reacts with hydrogen?

Answer 1

Ammonia is produced synthetically by the reaction below. How many moles of $\mathrm{NH}_3$ are formed when $200.0\mathrm{g}$ of $\mathrm{N}_2$ reacts with hydrogen?

Solution:

Ammonia is produced by the reaction:

$$\mathrm{N}_2 + 3\mathrm{H}_2 = 2\mathrm{NH}_3.$$

Bases on this reaction, of 1 mole $\mathrm{N}_2$ create 2 moles $\mathrm{NH}_3$.

We'll determine how many moles are in $200.0\mathrm{g}$ of $\mathrm{N}_2$:

$$v(\mathrm{N}_2) = m(\mathrm{N}_2)/M(\mathrm{N}_2) = 200.0 / 28.0 = 7.14 \text{ mole}.$$

Consequently, from 7.14 moles $\mathrm{N}_2$ formed 2·7.14=14.28 moles of $\mathrm{NH}_3$.

Answer:

Quantity moles of $\mathrm{NH}_3$ are 14.28.