Question

Consider the following reaction between sulfur trioxide and water:

Answer 1

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%