Question

question 2
what mass of n₂o₄ (g) is needed to produce 48.5 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 \(\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%\).
We know that the actual yield of \(N_2\) is \(m = 48.5\space g\) and the percent yield is \(60\%\).
Let the theoretical yield of \(N_2\) be \(m_{theo}\). Then \(60\%=\frac{48.5}{m_{theo}}\times 100\%\), so \(m_{theo}=\frac{48.5\times100}{60}\space g\approx80.83\space g\).

Step2: Use the stoichiometry of the reaction

The balanced chemical equation is \(2N_2H_4(g)+N_2O_4(g)\to3N_2(g)+4H_2O(g)\).
The molar mass of \(N_2\) is \(M_{N_2}=2\times14.01\space g/mol = 28.02\space g/mol\), and the molar mass of \(N_2O_4\) is \(M_{N_2O_4}=2\times14.01 + 4\times16.00\space g/mol=92.02\space g/mol\).
From the balanced equation, the mole ratio of \(N_2O_4\) to \(N_2\) is \(n_{N_2O_4}:n_{N_2}=1:3\).
The number of moles of \(N_2\) (theoretical) is \(n_{N_2}=\frac{m_{theo}}{M_{N_2}}=\frac{80.83}{28.02}\space mol\approx2.885\space mol\).
Since \(n_{N_2O_4}=\frac{1}{3}n_{N_2}\), then \(n_{N_2O_4}=\frac{2.885}{3}\space mol\approx0.9617\space mol\).
The mass of \(N_2O_4\) is \(m_{N_2O_4}=n_{N_2O_4}\times M_{N_2O_4}\).
Substitute \(n_{N_2O_4}\approx0.9617\space mol\) and \(M_{N_2O_4} = 92.02\space g/mol\) into the formula: \(m_{N_2O_4}=0.9617\times92.02\space g\approx88.50\space g\).

Answer:

\(88.50\)