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what is the percent yield for the reaction below when 364 g so, and 420 g
o2 produce 408 g s03?
2502(9) + o2(g) → 2803(9
o a. 51.5%
ob. 89.7%
oc. 97.1%
o d. 100%

Percent yield = 196%

Explanation:

Given data;

Mass of SO₂ = 364 g

Mass of O₂ = 42.0 g

Mass of SO₃ = 408 g

Percent yield = ?

Solution:

Chemical equation:

2SO₂ + O₂ → 2SO₃

Number of moles of SO₂:

Number of moles = mass/ molar mass

Number of moles = 364 g / 64.066 g/mol

Number of moles = 5.7 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 42.0 g / 32 g/mol

Number of moles = 1.3 mol

Now we compare the moles ammonia with hydrogen and nitrogen

                             O₂              :      SO₃

                               1               :        2

                               1.3            :      2/1 ×1.3 = 2.6 mol                                 

                             SO₂           :        SO₃

                                 2            :        2

                                 5.7          :      5.7

The number of moles of  SO₃ produced by oxygen are less so it will limiting reactant.

Theoretical yield:

Mass = number of moles × molar mass

Mass = 2.6 mol × 80.07 g/mol

Mass = 208.2 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 408 g / 208.2 g × 100

Percent yield = 1.96 × 100

Percent yield = 196%