Question

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N204 and 45.0 g N2H4. Some possibly useful molar masses are as follows: N2O4 92.02 g/mol, N2H4 32.05 g/mol N204) 2 N2H4(1)3 N2(g) + 4 H2O(g)

Answer 1

Answer: The mass of nitrogen gas produced will be 45.64 grams.

Step-by-step explanation:

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

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

• For
  `N_2O_4`

Given mass of
`N_2O_4=50.0g`

Molar mass of
`N_2O_4=92.02g/mol`

Putting values in equation 1, we get:

`\text{Moles of }N_2O_4=(50g)/(92.02g/mol)=0.543mol`

• For
  `N_2H_4`

Given mass of
`N_2H_4=45.0g`

Molar mass of
`N_2O_4=32.05g/mol`

Putting values in equation 1, we get:

`\text{Moles of }N_2H_4=(45g)/(32.05g/mol)=1.40mol`

For the given chemical reaction:

`N_2O_4(l)+2N_2H_4(l)\rightarrow 3N_2(g)+4H_2O(g)`

By stoichiometry of the reaction:

1 mole of
`N_2O_4` reacts with 2 moles of
`N_2H_4`

So, 0.543 moles of
`N_2O_4` will react with =
`(2)/(1)* 0.543=1.086moles` of
`N_2H_4`

As, the given amount of
`N_2H_4` is more than the required amount. Thus, it is considered as an excess reagent.

Hence,
`N_2O_4` is the limiting reagent.

By Stoichiometry of the reaction:

1 mole of
`N_2O_4` produces 3 moles of nitrogen gas

So, 0.543 moles of
`N_2O_4` will produce =
`(3)/(1)* 0.543=1.629moles` of nitrogen gas.

Now, calculating the mass of nitrogen gas from equation 1, we get:

Molar mass of nitrogen gas = 28.02 g/mol

Moles of nitrogen gas = 1.629 moles

Putting values in equation 1, we get:

`1.629mol=\frac{\text{Mass of nitrogen gas}}{28.02g/mol}\\\\\text{Mass of nitrogen gas}=45.64g`

Hence, the mass of nitrogen gas produced will be 45.64 grams.