## Answers

8 SO_2 + 16 H_2 S rightarrow 3 S_8 + 16 H_2 O moles of SO_2 = 93.0/64 Rightarrow 1.45 moles moles of H_2 S = 93.0/34 Rightarrow 2.73 moles 8 moles of SO_2 gives rightarrow 3 moles of S_8 1.45 moles of SO_2 gives rightarrow 3 times 1.45/8 Rightarrow 0.5437 moles of S_8 16 moles of H_2 S gives rightarrow 3 moles of S_8 2.73 moles of H_2 S gives rightarrow 3 times 2.73/16 Rightarrow 0.5118 moles of S_8 Limiting reagent is H_2 S maximum mass of S_8 formed = 0.5118 times 256 Rightarrow 131.02 g of S_8 N_2 (g) + 3 H_2 (g) rightarrow 2 NH_3 (g) 1 molecule of N_2 requires rightarrow 3 molecules of H_2 H_2 is the limiting reagent molecules of NH_3 formed = 6 molecules of NH_3 H_2 molecules remain = zero N_2 molecules remain = 1 molecule H_2 is the limiting reagent Fe_2 O_3 + 3 C rightarrow 2 Fe + 3 CO 1 mole of Fe O_3 gives rightarrow 2 moles of Fe 0.387 moles of Fe_2 O_3 gives rightarrow 2 times 0.387/1 Rightarrow 0.774 moles of Fe mass of 0.774 moles of Fe = 0.774 times 55.8 Rightarrow 43.18 g of Fe Actual yield = 18.3 of Fe % yield = 18.3/43.18 times 100 Rightarrow 42.38 %

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