Question

question 4
what mass of n₂o₄(g) is needed to produce 42.85 g of n₂(g) if the percent yield is 60%? round your answer to two decimal places and do not include units.
2n₂h₄(g) + n₂o₄(g) → 3n₂(g) + 4h₂o(g)
add your answer
1 point

Explanation:

Step1: Calculate the theoretical yield of \(N_2\)

The percent - yield formula is \(\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%\). Given \(\text{Percent Yield} = 60\%\) and \(\text{Actual Yield}=42.85\) g. Let the theoretical yield of \(N_2\) be \(x\). Then \(60=\frac{42.85}{x}\times100\), so \(x=\frac{42.85\times100}{60}\approx71.42\) g.

Step2: Use stoichiometry

From the balanced equation \(2N_2H_4(g)+N_2O_4(g)\to3N_2(g)+4H_2O(g)\), the molar ratio of \(N_2O_4\) to \(N_2\) is \(1:3\). The molar mass of \(N_2\) is \(M_{N_2}=2\times14 = 28\) g/mol, and the molar mass of \(N_2O_4\) is \(M_{N_2O_4}=2\times14 + 4\times16=92\) g/mol.
If \(n_{N_2}=\frac{\text{Theoretical Yield of }N_2}{M_{N_2}}=\frac{71.42}{28}\) mol. Then \(n_{N_2O_4}=\frac{1}{3}n_{N_2}\).
The mass of \(N_2O_4\), \(m_{N_2O_4}=n_{N_2O_4}\times M_{N_2O_4}\). \(n_{N_2}=\frac{71.42}{28}\approx2.55\) mol. \(n_{N_2O_4}=\frac{2.55}{3}=0.85\) mol. \(m_{N_2O_4}=0.85\times92 = 78.2\) g.

Answer:

\(78.20\)