Question

consider the following reaction: 2 clf₃ + 2 nh₃ → n₂ + 6 hf + cl₂. if 27.9 g of nh₃ and 165.8 g of clf₃ are reacted together, assuming complete reaction, how much cl₂ would be formed, in g?

Explanation:

Step1: Calculate the molar mass of \(NH_3\) and \(ClF_3\)

The molar mass of \(NH_3\) (\(M_{NH_3}\)): \(M_{NH_3}=14 + 3\times1=17\space g/mol\)
The molar mass of \(ClF_3\) (\(M_{ClF_3}\)): \(M_{ClF_3}=35.5+3\times19 = 92.5\space g/mol\)
The molar mass of \(Cl_2\) (\(M_{Cl_2}\)): \(M_{Cl_2}=2\times35.5 = 71\space g/mol\)

Step2: Calculate the number of moles of \(NH_3\) and \(ClF_3\)

The number of moles of \(NH_3\) (\(n_{NH_3}\)): \(n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{27.9\space g}{17\space g/mol}\approx1.64\space mol\)
The number of moles of \(ClF_3\) (\(n_{ClF_3}\)): \(n_{ClF_3}=\frac{m_{ClF_3}}{M_{ClF_3}}=\frac{165.8\space g}{92.5\space g/mol}\approx1.79\space mol\)

Step3: Determine the limiting reactant

From the balanced equation \(2ClF_3 + 2NH_3
ightarrow N_2+6HF + Cl_2\), the mole ratio of \(ClF_3\) to \(NH_3\) is \(1:1\).
Since \(n_{NH_3}

Step4: Calculate the number of moles of \(Cl_2\) formed

From the balanced equation, the mole ratio of \(NH_3\) to \(Cl_2\) is \(2:1\).
If \(n_{NH_3} = 1.64\space mol\), then \(n_{Cl_2}=\frac{1}{2}n_{NH_3}=\frac{1}{2}\times1.64\space mol = 0.82\space mol\)

Step5: Calculate the mass of \(Cl_2\) formed

\(m_{Cl_2}=n_{Cl_2}\times M_{Cl_2}=0.82\space mol\times71\space g/mol = 58.22\space g\)

Answer:

58.22 g