Question

Hydrazine, N2H4 , reacts with oxygen to form nitrogen gas and water. N2H4(aq)+O2(g)⟶N2(g)+2H2O(l) If 2.65 g of N2H4 reacts with excess oxygen and produces 0.350 L of N2 , at 295 K and 1.00 atm, what is the percent yield of the reaction?

Answer 1

Answer: The percent yield of the reaction is 17.41 %.

Explanation:

• For $N_2H_4$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

Given mass of $N_2H_4$ = 2.65 g

Molar mass of $N_2H_4$ = 32.04 g/mol

Putting values in above equation, we get:

$\text{Moles of }N_2H_4=\frac{2.65g}{32.04g/mol}=0.0827mol$

• For $N_2$

To calculate the number of moles, we use the equation given by ideal gas equation:

$PV=nRT$

where,

P = pressure of the gas = 1 atm

V = Volume of gas = 0.350 L

n = Number of moles = ?

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = 273 K

Putting values in above equation, we get:

$1.00atm\times 0.350L=n\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 295K\\\\n=0.0144mol$

Now, to calculate the experimental yield of $N_2$, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$   .....(1)

We are given:

Moles of nitrogen gas = 0.0144 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

$0.0144mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=0.4032g$

Experimental yield of nitrogen gas = 0.4032 g

• For the given chemical equation:

$N_2H_4(aq.)+O_2(g)\rightarrow N_2(g)+2H_2O(l)$

By Stoichiometry of the reaction:

1 mole of $N_2H_4$ produces 1 mole of nitrogen gas.

So, 0.0827 moles of $N_2H_4$ will produce = $\frac{1}{1}\times 0.0827=0.0827mol$ of nitrogen gas.

Now, to calculate the theoretical yield of nitrogen gas, we use equation 1:

Moles of nitrogen gas = 0.0827 mol

Molar mass  nitrogen gas = 28 g/mol

Putting values in above equation, we get:

$0.0827mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=2.3156g$

Theoretical yield of nitrogen gas = 2.3156 g

• To calculate the percentage yield of nitrogen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of nitrogen gas = 0.4032 g

Theoretical yield of nitrogen gas = 2.3156 g

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{0.4032g}{2.3156g}\times 100\\\\\% \text{yield of nitrogen gas}=17.41\%$

Hence, the percent yield of the reaction is 17.41 %.

Answer 2

When 2.65 g of N₂H₄ reacts with excess oxygen and produces 0.350 L of N₂, at 295 K and 1.00 atm, the percent yield of the reaction is 17.5 %.

Let's consider the following balanced equation.

N₂H₄(aq) + O₂(g) ⟶ N₂(g) + 2 H₂O(l)

We can calculate the theoretical volume of N₂ obtained at STP from 2.65 g of N₂H₄ considering the following conversion factors.

• The molar mass of N₂H₄ is 32.05 g/mol.
• The molar ratio of N₂H₄ to N₂ is 1:1.
• At STP, 1 mole of N₂ occupies 22.4 L

$2.65 g N_2H_4 \times \frac{1molN_2H_4}{32.05gN_2H_4} \times \frac{1molN_2}{1molN_2H_4} \times \frac{22.4LN_2}{1molN_2} = 1.85 L$

We have 1.85 L of N₂ at 273.15 K, we can calculate the volume at 295 K using Charles' law.

$\frac{V_1}{T_1} = \frac{V_2}{T_2} \\\\V_2 = \frac{V_1 \times T_2}{T_1} = \frac{1.85L \times 295K}{273.15K} = 2.00 L$

Finally, we can calculate the percent yield using the following expression.

$\% yield = \frac{experimental\ yield}{theoretical\ yield} \times 100\% = \frac{0.350L}{2.00L} \times 100\% = 17.5\%$

When 2.65 g of N₂H₄ reacts with excess oxygen and produces 0.350 L of N₂, at 295 K and 1.00 atm, the percent yield of the reaction is 17.5 %.

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