Question

question 17
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 hf would be formed, in g?
1 pts

Explanation:

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

The molar mass of \(NH_3\) (\(M_{NH_3}\)): \(M_{NH_3}=14+(1\times3)=17\space g/mol\)
The molar mass of \(ClF_3\) (\(M_{ClF_3}\)): \(M_{ClF_3}=35.5+(19\times3)=35.5 + 57=92.5\space g/mol\)
The molar mass of \(HF\) (\(M_{HF}\)): \(M_{HF}=1 + 19=20\space g/mol\)

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 chemical equation \(2ClF_3+2NH_3
ightarrow N_2 + 6HF+Cl_2\), the mole ratio of \(ClF_3\) to \(NH_3\) is \(1:1\).
Since \(n_{NH_3}\approx1.64\space mol\) and \(n_{ClF_3}\approx1.79\space mol\), \(NH_3\) is the limiting reactant.

Step4: Calculate the number of moles of \(HF\) produced

According to the mole ratio in the balanced equation, the mole ratio of \(NH_3\) to \(HF\) is \(2:6 = 1:3\).
If \(n_{NH_3} = 1.64\space mol\), then \(n_{HF}=3\times n_{NH_3}\) (because for every \(2\) moles of \(NH_3\), \(6\) moles of \(HF\) are produced)
\(n_{HF}=3\times1.64\space mol = 4.92\space mol\)

Step5: Calculate the mass of \(HF\) produced

Using the formula \(m = n\times M\), where \(n = n_{HF}\) and \(M = M_{HF}\)
\(m_{HF}=n_{HF}\times M_{HF}=4.92\space mol\times20\space g/mol = 98.4\space g\)

Answer:

\(98.4\space g\)