According to the following reaction, how many grams of water are required for the complete reaction of 29.0 grams of sulfur dioxide?
sulfur dioxide (g) + water (l) sulfurous acid (H2SO3) (g)
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Expert's answer
2018-10-03T05:04:16-0400
m(SO2) = 29 g M(SO2) = 32 + 16*2 = 64 g/mol n(SO2) = m/M = 29g/64g/mol = 0.453 mol According to the following reaction 1 mol of sulfur dioxide reacts with 1 mol of water. Then 0.453 mol of water are required for the complete reaction of 0.453 mol of sulfur dioxide. n(H2O) = 0.453 mol M(H2O) = 18 g/mol m(H2O) = n*M = 0.453mol*18g/mol = 8.154 g
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