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 $NH_3$.

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

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 $\frac{19.12}{2}=9.56$ moles of $N_2$

Now we have to calculate the mass of $N_2$.

$\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2$

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

Therefore, the mass of nitrogen needed are, 267.68 grams