Question

Sulfur dioxide gas (SO2) reacts with excess oxygen gas (O2) and excess liquid water (H2O) to form liquid sulfuric acid (H2SO4). In the laboratory, a chemist carries out this reaction with 67.2 L of sulfur dioxide and gets 250 g of sulfuric acid.

• Write a balanced equation for the reaction.
• Calculate the theoretical yield of sulfuric acid.
• Calculate the percent yield of the reaction. (One mole of any gas occupies 22.4 L under certain conditions of temperature and pressure. Assume those conditions for this question.)

Answer 1

Part A

The balanced equation is as below

2SO₂ +O₂ +2H₂O → 2 H₂SO₄

part B

The theoretical yield of H₂SO₄ is 294 grams

calculation

step 1 :find the moles of SO₂

That is at STP 1 moles = 22.4 L

? moles = 67.2 L

by cross multiplication

= (67.2 L x 1 moles) / 22.4 L = 3 moles

step 2: use the mole ratio to determine the moles of H₂SO₄

from equation above the mole ratio of SO₂:H₂SO₄ is 2:2 = 1:1 therefore the moles of H₂SO₄ is also 3 moles

step 3: find the theoretical yield

theoretical yield= moles x molar mass

from periodic table the molar mass of H₂SO₄ = (1 x2) + 32 + (16 x4) =98 g/mol

Theoretical yield = 3 moles x 98 g/mol = 294 grams

part c

percent yield = actual yield/ theoretical yield x 100

=( 250g / 294 g) x 100 = 85.03%