Answer:
A. H2O, B. 60.9 g, C. 86.3%
Explanation:
SO3 (g) + H2O (l) → H2SO4 (aq)
Chemist equation is ballanced: ok
1 mol of SO3 reacts with 1 mol of water, to produce 1 mol of H2SO4
First step: Find out the mol of each reactant.
Mass of SO3 = 80 g/m
Mol of SO3 → mass /molar weight = 61.5 g /80g/m = 0.768 mol
Mass of water = 18 g/m
Mol of H2O → mass /molar weight = 11.2 g /18g/m = 0.622 mol
Second step: Discover the limiting reactant.
As the relation is 1:1 (SO3 - H2O), 0.768 mol of SO3 needs 0.768 mol of water and 0.622 mol of water needs 0.622 mol of SO3.
I don't have enough water, so H2O is my limiting.
Third step: Find out the yield.
Now that we know the limiting reactant we can work, and relation between products is still 1:1 so 0.622 mol of water, produce 0.622 mol of H2SO4
Molar weight of H2SO4 = 98 g/m
Molar weight . mol = mass → 98 g/m . 0.622 mol = 60.9 g
This is the 100% yield (Theoretical one) but the chemist collects only 52.6 g of sulfuric so to find the real yield, we can use the rule of three.
60.9 g _____ 100 %
52.6 g ______ (52.6 .100) /60.9 0 = 86.3%
