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

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

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Alleei Alleei · Answer 2 · 2019-07-06T00:39:17-04:00

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