Question

1. how many moles of nh₃ are required to produce 10.0 mol of h₂o? 4nh₃(g) + 7o₂(g) → 4no₂(g) + 6h₂o(g) a. 3.78 mol b. 5.44 mol c. 6.67 mol d. 1.23 mol e. 2.89 mol

Explanation:

Step1: Determine mole ratio

From the balanced equation \(4NH_3(g)+7O_2(g)\to4NO_2(g)+6H_2O(g)\), the mole ratio of \(NH_3\) to \(H_2O\) is \(\frac{4}{6}=\frac{2}{3}\).

Step2: Calculate moles of \(NH_3\)

Let \(n(NH_3)\) be the moles of \(NH_3\). We know \(n(H_2O) = 10.0\space mol\). Using the mole - ratio \(\frac{n(NH_3)}{n(H_2O)}=\frac{2}{3}\), so \(n(NH_3)=\frac{2}{3}\times n(H_2O)\). Substitute \(n(H_2O) = 10.0\space mol\) into the equation: \(n(NH_3)=\frac{2\times10.0}{3}=\frac{20}{3}\approx6.67\space mol\).

Answer:

C. \(6.67\space mol\)