If 7.0 mol of no and 5.0 mol of o2 are reacted tegethor. the reaction generates 3.0 mol of no2. what is the percent yield for the reaction? 2no(g) + o2(g) -> 2no2(g)

Percentage yield = 30%

Explanation:

Given data:

Number of moles of NO = 7.0 mol

Number of moles of O₂ = 5 mol

Number of moles of NO₂ = 3 mol

Percentage yield = ?

Solution:

Chemical equation:

2NO + O₂ → 2NO₂

Now we will compare the moles of NO₂ with NO and O₂ .

                  NO           :               NO₂

                  2               :               2

                 7.0             :              7.0

                O₂               :                NO₂

                 1                 :                 2

                 5.0             :               2 ×5.0 = 10 mol

The number of moles of NO₂ produced by NO are less it will be limiting reactant.

Mass of NO₂ = moles × molar mass

Mass of NO₂ = 10 mol × 46g/mol

Mass of NO₂ =  460 g

Actual yield of NO₂:

Mass of NO₂ = moles × molar mass

Mass of NO₂ = 3 mol × 46g/mol

Mass of NO₂ =  138 g

Percentage yield:

Percentage yield = Actual yield/theoretical yield × 100

Percentage yield = 138 g/ 460 g × 100

Percentage yield = 30%

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