Question

According to the following reaction, how many grams of sulfurous acid (H₂SO₃) will be formed upon the complete reaction of 30.0 grams of water with excess sulfur dioxide?

SO₂ (g) + H₂O (l) =H₂SO₃ (g)

grams sulfurous acid (H₂SO₃)

Answer 1

the given Reaction will be Soz (g) + H2O ce = H2 503 g) as the equation is balanced and so 2 s in excess. So H₂O is the limiting reagent. no of moles of H2O = weight of H2O Molor mais al H2O - 30 1.66 moles 13 I mole of H₂O gives a moles of H ₂ SO 3 so to according the unitare method og 1.66 mole H2O will give di 66 mole of H₂ 503 the g mass col H₂SO3 in gram will be Molar mass of M₂ SO3 = 82.07 = 1.66 x 82.07 = "136.23 grams