Question

2 N2H4(g)+ N2O4(g) à3 N2(g) + 4 H2O(g) When 8.0 g of N2H4(32 g mol-1) and 18.4 g of N2O4(92 g mol-1) are mixed together and react according to the equation above, what is the maximum mass of H2O that can be produced?

Answer 1

Answer: The mass of water that can be produced is 9.22 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_2H_4` :

Given mass of
`N_2H_4` = 8.2 g

Molar mass of
`N_2H_4` = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of }N_2H_4=(8.2g)/(32g/mol)=0.256mol`

• For
  `N_2O_4` :

Given mass of
`N_2O_4` = 18.4 g

Molar mass of
`N_2O_4` = 92 g/mol

Putting values in equation 1, we get:

`\text{Moles of }N_2O_4=(18.4g)/(92g/mol)=0.2mol`

The given chemical equation follows:

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

By Stoichiometry of the reaction:

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

So, 0.256 moles of
`N_2H_4` will react with =
`(1)/(2)* 0.256=0.128mol` of
`N_2O_4`

As, given amount of
`N_2O_4` is more than the required amount. So, it is considered as an excess reagent.

Thus,
`N_2H_4` is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of
`N_2H_4` produces 4 moles of water

So, 0.256 moles of
`N_2H_4` will produce =
`(4)/(2)* 0.256=0.512moles` of water

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

Molar mass of water = 18 g/mol

Moles of water = 0.512 moles

Putting values in equation 1, we get:

`0.512mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.512mol* 18g/mol)=9.22g`

Hence, the mass of water that can be produced is 9.22 grams

Answer 2

9.0 g

Step-by-step explanation:

First of all, we need to find which of the reagents is the limiting, that is, the one that will be fulled consumed during the reaction. For the balanced reaction, 2 mols of N2H4 reacts with 1 mol of N2O4, so the stoichiometry is 2:1.

The first step is always to use the information in mol and then put the information in the unity required. In this case, the question wants the mass, so we must multiply the molar mass by the stoichiometry coefficients, and then use one of the mass to test and find which of the reagents is the limiting. Let's use the mass of N2H4 as a test:

2x 32 g/mol of N2H4 -------------------- 1x 92 g/mol of N2O4

8 g of N2H4 --------------------------------- X

By a direct rule of three:

X = 11.5 g of N2O4

If we use 8 g of N2H4 we will need 11.5 g of N2O4, and it was added 18.4 g, so it's in excess and N2H4 is the limiting reagent.

Now, we do the relation between the limiting reagent and H2O, which has molar mass equal to 18 g/mol ( 2x 1 g/mol of H + 16 g/mol of O). Now the stoichiometric is 2:4, or 1:2

1x 32 g/mol of N2H4 --------------------- 2x 18 g/mol of H2O

8 g of N2H4 --------------------------------- Y

By a direct rule of three:

Y = 9.0 g of H2O.