Question

56g of n2 reacts with 9g of h2 in a closed bessel to form nh3. which reactant is left in excess and how much​

Answer 1

Explanation:

3H2+N2=NH3

n(H2)=m/M

=9/2

4.5 mol

n(N2)=m/M

56/28

=2 mol

:. n(H2)/3 and n(N2)/1

4.5/3 and 2/1

1.5 and 2

:. N2 is an excess reagent since it has a higher coeficient ratio....

Answer 2

Answer:

14g or 0.5 moles of nitrogen is left in excess.

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EXPLANATION :

N2 + 3H2 —> 2NH3

ie, 1 mole (2*14 = 28g) of nitrogen reacts with 3 moles (3*1*2=6g) of hydrogen to give 2moles (2*(14+3)=34g) of ammonia.

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given, 56g of N2 reacts with 9g of h2 in a closed vessel to form nh3.

if 6g of H2 react with 28g of N2 to give 34g of NH3,

then 1g of H2 will react with 28/6g of N2 to give 34/6g of NH3.

therefore,

9g of H2 will react with 28/6 ×9 = 42g of N2

to give 34/6 ×9 = 51g of NH3.

as you can see, hydrogen is the limiting reagent here. (That is, it is the reactant that is being completely used up in the reaction, and preventing any further reaction from occurring.)

And nitrogen is left in excess.

Excess nitrogen left is 56-42  = 14g

= 14/28 = 0.5 moles.