Question

ammonia reacts with diatomic oxygen to form nitric oxide and water vapor. what is the theoretical yield of water, in moles, when 40.0 g of ammonia and 50.0 g of oxygen are mixed and allowed to react? 4 nh₃(g) + 5 o₂(g) → 4 no(g) + 6 h₂o(g) 1.30 mol 2.54 mol 1.57 mol 3.52 mol 1.88 mol

Explanation:

Step1: Calculate moles of reactants

• Molar mass of \(NH_3\): \(M_{NH_3}=14 + 3\times1=17\space g/mol\). Moles of \(NH_3\), \(n_{NH_3}=\frac{40.0\space g}{17\space g/mol}\approx2.35\space mol\)
• Molar mass of \(O_2\): \(M_{O_2} = 32\space g/mol\). Moles of \(O_2\), \(n_{O_2}=\frac{50.0\space g}{32\space g/mol}\approx1.56\space mol\)

Step2: Determine limiting reactant

From the balanced equation \(4NH_3(g)+5O_2(g)\to4NO(g)+6H_2O(g)\), the mole ratio of \(NH_3\) to \(O_2\) is \(\frac{n_{NH_3}}{n_{O_2}}=\frac{4}{5}\).
If \(NH_3\) is the limiting reactant, moles of \(O_2\) required \(n_{O_2}^{req}=\frac{5}{4}\times n_{NH_3}=\frac{5}{4}\times2.35\space mol\approx2.94\space mol\). But we have only \(1.56\space mol\) of \(O_2\).
If \(O_2\) is the limiting reactant, moles of \(NH_3\) required \(n_{NH_3}^{req}=\frac{4}{5}\times n_{O_2}=\frac{4}{5}\times1.56\space mol = 1.25\space mol\). We have \(2.35\space mol\) of \(NH_3\) (excess). So \(O_2\) is the limiting reactant.

Step3: Calculate moles of \(H_2O\)

From the balanced equation, mole ratio of \(O_2\) to \(H_2O\) is \(\frac{n_{O_2}}{n_{H_2O}}=\frac{5}{6}\).
\(n_{H_2O}=\frac{6}{5}\times n_{O_2}\)
Substitute \(n_{O_2} = 1.56\space mol\)
\(n_{H_2O}=\frac{6}{5}\times1.56\space mol=1.87\space mol\approx1.88\space mol\)

Answer:

1.88 mol