Questions: Question 22 of 40 How many moles of NH3 can be produced from 4.01 moles of nitrogen in the following reaction: N2(g) + 3 H2(g) → 2 NH3(g) 1 2 4 5 7 8

Solution Steps

We are given 4.01 moles of nitrogen (\(\mathrm{N}_2\)) and need to determine how many moles of ammonia (\(\mathrm{NH}_3\)) can be produced.

The balanced chemical equation for the reaction is: \[ \mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g}) \]

From the balanced equation, 1 mole of \(\mathrm{N}_2\) produces 2 moles of \(\mathrm{NH}_3\).

Using the mole ratio, we can calculate the moles of \(\mathrm{NH}_3\) produced from 4.01 moles of \(\mathrm{N}_2\): \[ \text{Moles of } \mathrm{NH}_3 = 4.01 \text{ moles of } \mathrm{N}_2 \times \frac{2 \text{ moles of } \mathrm{NH}_3}{1 \text{ mole of } \mathrm{N}_2} = 8.02 \text{ moles of } \mathrm{NH}_3 \]

Final Answer