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Question:

For a reaction N2(g) + 3H2(g) → 2NH3(g) identify dihydrogen (H2) as a limiting reagent in the following reaction mixture: A) 28g of N2 + 6g of H2 B) 35g of N2 + 8g of H2 C) 14g of N2 + 4g of H2 D) 56g of N2 + 10g of H2

28gofN2+6gofH2

35gofN2+8gofH2

14gofN2+4gofH2

56gofN2+10gofH2

Solution:

Correct option is C. 56g of N2 + 10g of H2
N2(g) + 3H2(g) → 2NH3(g)
In the given reaction, 1 mol of N2 requires 3 moles of H2 for the formation of ammonia. Thus, the number of moles of N2 and H2 should be in the ratio 1:3.
We know, n = weight / molecular weight
Thus in option C, 56g of N2 means 2 moles of N2 and 10g of H2 means 5 moles of H2. Thus here H2 acts as a limiting reagent.