Question

Sulfur trioxide dissolves in water, producing H2SO4. How much sulfuric acid can be produced from 10.1 mL of water (d= 1.00 g/mL) and 23.9 g of SO3? How much of the reagent in excess is left over?

Answer 1

29.2 g of H₂SO₄ are produced

0.263 moles of water remain after the reaction goes complete.

Step-by-step explanation:

We make the reaction in the first step:

Reactants are water and SO₃

H₂O + SO₃ → H₂SO₄

Let's determine moles of reactants:

23.9 g . 1 mol / 80.06g = 0.298 moles

We apply density, to determine mass of water:

D = m/ V so m = D . V

1 g/mL . 10.1 mL = 10.1 g

moles of water are: 10.1 g . 1 mol/ 18g = 0.561 moles

As ratio is 1:1, for 0.298 moles of SO₃ we need the same amount of water, and we have 0.561 moles. Then, water is the excess reagent and sulfur trioxide is the limiting.

0.561 - 0.298 = 0.263 moles of water that remain after the reaction goes complete.

As ratio is 1:1, again, 0.298 moles of SO₃ can produce 0.298 moles of acid.

We determine the mass: 0.298 mol . 98.06 g /mol = 29.2 g