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Archy [21]
3 years ago
7

When 28 g of nitrogen and 6 g of hydrogen react, 34 g of ammonia are produced. if 100 g of nitrogen react with 6 g of hydrogen,

how much ammonia will be produced?
Chemistry
2 answers:
nignag [31]3 years ago
8 0

Answer : The mass of NH_3 produced will be, 34 grams.

Explanation : Given,

Mass of N_2 = 100 g

Mass of H_2 = 6 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{100g}{28g/mole}=3.57moles

\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{6g}{2g/mole}=3moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 3 mole of H_2 react with 1 mole of N_2

So, the given 3 moles of N_2 react with 1 moles of H_2

From this we conclude that, N_2 is an excess reagent because the given moles are greater than the required moles and H_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3.

As, 3 moles of H_2 react to give 2 moles of NH_3

So, 3 moles of H_2 react to give \frac{3}{3}\times 2=2 moles of NH_3

Now we have to calculate the mass of NH_3.

\text{Mass of }NH_3=\text{Moles of }NH_3\times \text{Molar mass of }NH_3

\text{Mass of }NH_3=(2mole)\times (17g/mole)=34g

Therefore, the mass of NH_3 produced will be, 34 grams.

vampirchik [111]3 years ago
4 0
N2 + 3H2 --> 2NH3
When 100g of N2 , no of moles of N2= 100/(28)=3.57 mol
no. of moles of h2 = 6/(2)=3mol
Therefore h2 is limiting reagent.
no. of moles of ammonia= 3/3*2=2moles
mass of ammonia produced= 2 mol * (14+3)= 34g
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