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?

The percent yield of the reaction is 17.41 %.

Explanation:

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

Given mass of = 2.65 g

Molar mass of = 32.04 g/mol

Putting values in above equation, we get:

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

where,

P = pressure of the gas = 1 atm

V = Volume of gas = 0.350 L

n = Number of moles = ?

R = Gas constant =

T = temperature of the gas = 273 K

Putting values in above equation, we get:

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

  .....(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:

Experimental yield of nitrogen gas = 0.4032 g

By Stoichiometry of the reaction:

1 mole of produces 1 mole of nitrogen gas.

So, 0.0827 moles of will produce = 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:

Theoretical yield of nitrogen gas = 2.3156 g

Experimental yield of nitrogen gas = 0.4032 g

Theoretical yield of nitrogen gas = 2.3156 g

Putting values in above equation, we get:

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