A 1.75-L sample of H2S measured at 25.0°C and 625 mmHg is mixed with 5.75 L of O2, measured at 10.0 °C and 715 mmHg, and the mixture is allowed to react. 2 H2S(g) + 3 O2 (g) → 2 SO2 (g) + 2 H2O (g) How much H2O in grams, is produced?
n(H2S) = V/Vm = 1.75L/22.4L/mol = 0.078 mol
n(O2) = 5.75L/22.4L/mol = 0.257 mol
H2S is limiting reactant, O2 is in the excess.
n(H2O) = 0.078 mol
m(H2O) = n*M = 0.078mol*18g/mol = 1.404 g produced
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