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Answer 1

Hii
Nice question

N2 + 3H2 --- 2NH3

From this equation , we can say that

1 mole of N2 reacts with 3 moles of H2

28g of N2 reacts with 6g of H2

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

Amount of H2 produced by 50 × 10³ g of N2 = 6/28 × 50 × 10³ g = 10.71 × 10³ g = 10.71 kg

But H2 is present in lesser amount than the required one.

So, H2 is the limiting reagent.

Now,
1 mole of H2 reacts with 2 moles of NH3

28g of H2 reacts with 34g of NH3

1g of H2 reacts with 34/28 g of NH3

Amount of NH3 produced by 10 × 10³ g of H2 = 34/28 × 10 × 10³ g = 12.14 × 10³ g = 12.14 kg

No. of moles of NH3 = given mass/ molar mass

No. of moles of NH3 = 12.14 × 10³ /17 = 0.71 × 10³ moles = 710 moles

Hope this helps you.