Question

Hydrazine, N2H4, may react with oxygen to form nitrogen gas and water.N2H4(aq) + O2(g) -> N2(g) + 2H2O(l)If 3.95 g of N2H4 reacts and produces 0.550 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 18.7 %

Explanation:

• Calculating the theoretical yield:

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

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

Given mass of hydrazine = 3.95 g

Molar mass of hydrazine = 32 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrazine}=\frac{3.95g}{32g/mol}=0.123mol$

The given chemical equation follows:

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

By Stoichiometry of the reaction:

1 mole of hydrazine produces 1 mole of nitrogen gas

So, 0.123 moles of hydrazine will produce = $\frac{1}{1}\times 0.123=0.123mol$ of nitrogen gas

• Calculating the experimental yield:

To calculate the moles of nitrogen gas gas, we use the equation given by ideal gas, which follows:

$PV=nRT$

where,

P = pressure of the nitrogen gas = 1.00 atm

V = Volume of the nitrogen gas = 0.550 L

T = Temperature of the nitrogen gas = 295 K

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of the nitrogen gas = ?

Putting values in above equation, we get:

$1.00atm\times 0.555L=n_{N_2}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 295K\\\\n_{N_2}=\frac{1.00\times 0.555}{0.0821\times 295}=0.023mol$

• Calculating the percentage yield:

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 = 0.023 moles

Theoretical yield of nitrogen gas = 0.123 moles

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{0.023}{0.123}\times 100\\\\\% \text{yield of nitrogen gas}=18.7\%$

Hence, the percent yield of the reaction is 18.7 %

Answer 2

Answer:

The percent yield of the reaction is 18.45 %

Explanation:

The reaction is:

N₂H₄(aq) + O₂(g) →  N₂(g) + 2H₂O(l)

Ratio between hydrazine and N₂ is 1:1, so 1 mol of hydrazine produces 1 mol of N₂

The molar mass of hydrazine is 32  g/m

The moles we used are : mass / molar mass hydrazine

3.95 g / 32 g/m = 0.123 moles

So, in the 100 % yield reaction, we will produce 0.123 of gas.

Let's apply the Ideal Gases law to find out, the moles of gas that have been produced.

0.550L . 1 atm = n . 0.082 . 295K

(0.550L .1atm) / 0.082 . 295K = n

0.0227 mol = n

So, to find out the percent yield of the reaction we finally make a rule of three

0.123 moles ___ 100 %

0.0227 moles ____ 18.45 %