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

A.2SO₂ + O₂ + 2H₂O → 2H₂SO₄
B.Moles of SO₂ = 67.2 / 22.4Moles of SO₂ = 3 molesMoles of H₂SO₄ = 3 molesMass of H₂SO₄ = 3 x 98Mass of H₂SO₄ = 294 grams
Assuming sulfuric acid to have the same density as water,density = 1000 g / LVolume = mass / densityVolume = 294 / 1000Volume = 0.29 liters of sulfuric acid

Answer 2

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%