Question

How many grams of N2 are required to produce 240.0g NH3?

Answer 1

Balanced equation: N2 + 3H2 ==> 2NH3
moles of NH3 in 240.0 g NH3: 240.0 g x 1 mol/17 g = 14.1 moles
moles N2 needed: 14.1 moles NH3 x 1 mole N2/2 moles NH3 = 7.05 moles N2 needed
grams N2 needed: 7.05 moles x 28 g/mole = 197.4 grams needed :)

Answer 2

Answer: The mass of
`N_2` required are, 198 grams.

Explanation : Given,

Mass of
`NH_3` = 240.0 g

Molar mass of
`NH_3` = 17 g/mol

First we have to calculate the moles of
`NH_3`.

`\text{Moles of }NH_3=\frac{\text{Given mass }NH_3}{\text{Molar mass }NH_3}`

`\text{Moles of }NH_3=(240.0g)/(17g/mol)=14.12mol`

Now we have to calculate the moles of
`N_2`

The balanced chemical equation is:

`N_2+3H_2\rightarrow 2NH_3`

From the reaction, we conclude that

As, 2 mole of
`NH_3` produces from 1 mole of
`N_2`

So, 14.12 mole of
`NH_3` produces form
`(14.12)/(2)=7.06` mole of
`N_2`

Now we have to calculate the mass of
`N_2`

`\text{ Mass of }N_2=\text{ Moles of }N_2* \text{ Molar mass of }N_2`

Molar mass of
`N_2` = 28 g/mole

`\text{ Mass of }N_2=(7.06moles)* (28g/mole)=197.68g\approx 198g`

Therefore, the mass of
`N_2` required are, 198 grams.