question 11\ndetermine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from…

Explanation:

Step1: Calculate molar masses

The molar mass of $N_2O_4$ is $M_{N_2O_4}=(2\times14 + 4\times16)\ g/mol=92\ g/mol$. The molar mass of $N_2H_4$ is $M_{N_2H_4}=(2\times14+4\times1)\ g/mol = 32\ g/mol$. The molar mass of $N_2$ is $M_{N_2}=2\times14\ g/mol = 28\ g/mol$.

Step2: Calculate moles of reactants

The number of moles of $N_2O_4$, $n_{N_2O_4}=\frac{m_{N_2O_4}}{M_{N_2O_4}}=\frac{50.0\ g}{92\ g/mol}\approx0.543\ mol$. The number of moles of $N_2H_4$, $n_{N_2H_4}=\frac{m_{N_2H_4}}{M_{N_2H_4}}=\frac{45.0\ g}{32\ g/mol}\approx1.406\ mol$.

Step3: Determine the limiting - reactant

From the balanced chemical equation $N_2O_4(l)+2N_2H_4(l)\rightarrow3N_2(g) + 4H_2O(g)$, the mole - ratio of $N_2O_4$ to $N_2H_4$ is $1:2$. For $0.543\ mol$ of $N_2O_4$, the moles of $N_2H_4$ required is $n_{N_2H_4\ required}=2\times n_{N_2O_4}=2\times0.543\ mol = 1.086\ mol$. Since the available moles of $N_2H_4$ ($1.406\ mol$) is more than the required moles ($1.086\ mol$), $N_2O_4$ is the limiting reactant.

Step4: Calculate moles of $N_2$ produced

From the balanced equation, the mole - ratio of $N_2O_4$ to $N_2$ is $1:3$. So the moles of $N_2$ produced, $n_{N_2}=3\times n_{N_2O_4}=3\times0.543\ mol = 1.629\ mol$.

Step5: Calculate mass of $N_2$ produced

The mass of $N_2$ produced, $m_{N_2}=n_{N_2}\times M_{N_2}=1.629\ mol\times28\ g/mol\approx45.6\ g$.

Answer:

Y N 45.6 g