Question

How many gram of nitrogen are needed to produce 325 gram of ammonia? n2(g) + 3h2(g) → 2nh3(g)

Answer 1

Molar mass :

N₂ = 28.0 g/mol
NH₃ = 17.0 g/mol

N₂(g) + 3 H₂(g) = 2 NH₃(g)

28.0 g (N₂)---------> 2 x 17 g NH₃
? g ------------> 325 g NH₃

Mass N₂ = ( 325 x 28.0 ) / ( 2 x 17 )

Mass N₂ = 9100 / 34

= 267.64 g of N₂

hope this helps!

Answer 2

Answer : The mass of nitrogen needed are, 267.68 grams

Explanation :

Mass of
`NH_3` = 325 g

Molar mass of
`NH_3` = 17 g/mole

Molar mass of
`N_2` = 28 g/mole

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

`\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}=(325g)/(17g/mole)=19.12moles`

Now we have to calculate the moles of
`N_2`.

The balanced chemical reaction is,

`N_2(g)+3H_2(g)\rightarrow 2NH_3(g)`

From the balanced reaction we conclude that

As, 2 moles of
`NH_3` obtained from 1 mole of
`N_2`

So, 19.12 moles of
`NH_3` obtained from
`(19.12)/(2)=9.56` moles 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`

`\text{Mass of }N_2=(9.56mole)* (28g/mole)=267.68g`

Therefore, the mass of nitrogen needed are, 267.68 grams