What is the percent yield for the reaction below when 705.0 g SO2 and 80.0 g 02 produce 586.0 g S03?
2802(g) + O2(g) → 2503(9)

Percent yield of reaction is 150%.

Explanation:

Given data:

Percent yield = ?

Actual yield of SO₃ = 586.0 g

Mass of SO₂ = 705.0 g

Mass of O₂ = 80.0 g

Solution:

Chemical equation:

2SO₂ + O₂      →     2SO₃

Number of moles of SO₂:

Number of moles = mass/ molar mass

Number of moles = 586.0 g/ 64.1 g/mol

Number of moles = 9.1 mol

Number of moles of O₂:

Number of moles = mass/ molar mass

Number of moles = 80.0 g/ 32g/mol

Number of moles = 2.5 mol

Now we will compare the mole of SO₃ with O₂ and SO₂.

                      SO₂          :          SO₃

                        2             :            2

                     9.1              :           9.1

                       O₂            :          SO₃

                        1              :            2

                     2.5             :           2×2.5 = 5

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

Theoretical yield of SO₃:

Mass = number of moles × molar mass

Mass = 5 mol × 80.1 g/mol

Mass = 400.5 g

Percent yield of reaction:

Percent yield = actual yield / theoretical yield  × 100

Percent yield = 586.0 g/ 400.5 g× 100

Percent yield = 1.5× 100

Percent yield = 150%

there are two significant figures.