Question

ammonia (nh₃) is a principal nitrogen fertilizer. it is prepared by the reaction between hydrogen (h₂) and nitrogen (n₂).
3h₂(g) + n₂(g) → 2nh₃(g)
in a particular reaction, 8.00 moles of nh₃ were produced. how many moles of h₂ and how many moles of n₂ were reacted to produce this amount of nh₃?
round each of your answers to 3 significant digits.
part 1 of 2
mol h₂
part 2 of 2
mol n₂

Explanation:

Step1: Determine the mole ratio

From the balanced chemical equation \(3H_2(g)+N_2(g)\to2NH_3(g)\), the mole ratio of \(H_2\) to \(NH_3\) is \(3:2\), and the mole ratio of \(N_2\) to \(NH_3\) is \(1:2\).

Step2: Calculate moles of \(H_2\)

Let \(n_{H_2}\) be the moles of \(H_2\). Using the mole ratio \(\frac{n_{H_2}}{n_{NH_3}}=\frac{3}{2}\). Given \(n_{NH_3} = 8.00\) mol. Then \(n_{H_2}=\frac{3}{2}\times n_{NH_3}\). Substitute \(n_{NH_3}\): \(n_{H_2}=\frac{3}{2}\times8.00 = 12.0\) mol.

Step3: Calculate moles of \(N_2\)

Let \(n_{N_2}\) be the moles of \(N_2\). Using the mole ratio \(\frac{n_{N_2}}{n_{NH_3}}=\frac{1}{2}\). Given \(n_{NH_3} = 8.00\) mol. Then \(n_{N_2}=\frac{1}{2}\times n_{NH_3}\). Substitute \(n_{NH_3}\): \(n_{N_2}=\frac{1}{2}\times8.00=4.00\) mol.

Answer:

\(12.0\) mol \(H_2\)
\(4.00\) mol \(N_2\)