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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?

User Colin Hale
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2 Answers

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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
User Wimnat
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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}=(100g)/(28g/mole)=3.57moles


\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=(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
(3)/(3)* 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* \text{Molar mass of }NH_3


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

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

User Badmadrad
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