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.)

The balanced equation is as below

2SO₂ +O₂ +2H₂O →  2 H₂SO₄

 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

percent yield =  actual yield/ theoretical yield x 100

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

the answer would be b i believe