28 g of N2 reacts with 10 g of H2 according to N2 + 3H2 -> 2NH3. The maximum mass of NH3 formed is (N = 14, H = 1)

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$28 g$ of $N X_{2}$ reacts with $10 g$ of $H X_{2}$ according to $N X_{2} + 3 H X_{2} 2 NH X_{3}$ . The maximum mass of $NH X_{3}$ formed is (N = 14, H = 1)

n(N2) = 1.00 mol; n(H2) = 5.0 mol. N2 needs only 3 mol H2, so N2 is limiting. NH3 = 2 x 1 = 2 mol = 34 g.

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