Question

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)

Answer 1

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