Question

50g of N2 gas and 10g of H2 gas are mixed to produce NH3 gas.Identify the limiting reagent in this reaction?​

Answer 1

N2 +3 H2 ---->2NH3

you can see that ,

1 mole of N2 react with 3mole of H2 form 2mole of NH3 .

e.g 28 g of N2 react with 6g of H2 form 34g of NH3

so,

1 g of N2 react with 6/28 g of H2

50g of N2 react with 6×50/28g of H2

6×50/28 g > 10g of H2

hence, it means H2 is limiting reagent because it is less then required amount .

now,

reaction proceed based on limiting reagents.

because , 6g of H2 form 34g NH3

so, 10g of H2 form 340/6 g of NH3

e.g mass of NH3 produced = 340/6 g

= 56.667 g

Answer 2

HELLO MATE【◆◆◆◆◆】【●●●●●】----+++++++++-------

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28 g of N2 react with 6 g of H2 to form 34 g of NH3

50 g of N2 will react with (6/28) x 50 = 10.71 g of H2

Available amount of H2 = 10 g

Therefore, H2 is the limiting reagent.

Now,

6 g H2 gives 34 g NH3

So, 10 g H2 will give (34/6 )x 10 = 56.67 g NH3.

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HOPE THIS HELP YOU...☺️☺️