Question

Can someone please help?

Answer 1

Answer: The correct option is, (a) 9.0 g

Explanation : Given,

Mass of
`N_2H_4` = 8.0 g

Mass of
`N_2O_4` = 92 g

Molar mass of
`N_2H_4` = 32 g/mol

Molar mass of
`N_2O_4` = 92 g/mol

First we have to calculate the moles of
`N_2H_4` and
`N_2O_4`.

`\text{Moles of }N_2H_4=\frac{\text{Given mass }N_2H_4}{\text{Molar mass }N_2H_4}`

`\text{Moles of }N_2H_4=(8.0g)/(32g/mol)=0.25mol`

and,

`\text{Moles of }N_2O_4=\frac{\text{Given mass }N_2O_4}{\text{Molar mass }N_2O_4}`

`\text{Moles of }N_2O_4=(92g)/(92g/mol)=1mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

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

From the balanced reaction we conclude that

As, 2 mole of
`N_2H_4` react with 1 mole of
`N_2O_4`

So, 0.25 moles of
`N_2H_4` react with
`(0.25)/(2)=0.125` moles of
`N_2O_4`

From this we conclude that,
`N_2O_4` is an excess reagent because the given moles are greater than the required moles and
`N_2H_4` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`H_2O`

From the reaction, we conclude that

As, 2 mole of
`N_2H_4` react to give 4 mole of
`H_2O`

So, 0.25 mole of
`N_2H_4` react to give
`(4)/(2)* 0.25=0.5` mole of
`H_2O`

Now we have to calculate the mass of
`H_2O`

`\text{ Mass of }H_2O=\text{ Moles of }H_2O* \text{ Molar mass of }H_2O`

Molar mass of
`H_2O` = 18 g/mole

`\text{ Mass of }H_2O=(0.5moles)* (18g/mole)=9.0g`

Therefore, the mass of
`H_2O` produced is, 9.0 grams.