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

User Lafeber
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2 Answers

4 votes
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!
User Alphaneo
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8.0k points
6 votes

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

User JerMah
by
8.4k points
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