Determine the percent yield for the reaction between 3.74 g of na and excess o2 if 5.34 g of na2o2 is recovered.

Answer is: the percent yield for the reaction is 84,2%.
Chemical reaction: 2Na + O₂ → Na₂O₂.
m(Na) = 3,74 g.
n(Na) = m(Na) ÷ M(Na).
n(Na) = 3,74 g ÷ 23 g/mol.
n(Na) = 0,162 mol.
From chemical reaction: n(Na) : n(Na₂O₂) = 2 : 1.
n(Na₂O₂) = 0,162 mol ÷ 2 = 0,081 mol.
m(Na₂O₂) = 0,081 mol · 78 g/mol.
m(Na₂O₂) = 6,341 g.
yield of reaction = 5,34 g / 6,341 g · 100%.
yield of reaction = 84,2%.

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mahima6824soni mahima6824soni · Answer 2 · 2022-03-04T06:28:18+01:00

The percentage yield for the reaction is 84.2%.

What is percent yield?

Percent yield is the difference in the percent of theoretical yield to the percent of actual yield.

The reaction is

[tex]\rm2Na + O_2 = Na_2O_2[/tex]

Mass of Na = 3.74g

Calculating the moles of Na

[tex]\rm Number\;of \;moles= \dfrac{mass}{molar\;mass}\\\\\\Number\;of \;moles= \dfrac{3.74}{23}= 0.162 \;mol[/tex]

The ratio of mole of Na : (Na₂O₂) = 2:1

[tex]\rm Number\;of \;moles= \dfrac{0.162}{2}= 0.081 \;mol\\\\mass\;of Na_2O_3 = 0.081 \times 78\;g/mol\\\\mass\;of Na_2O_3 = 6.341\;g\\\\\\Yield \;of\;reaction = \dfrac{5.34g}{6.341g} \times 100 = 84.2\%[/tex]

Thus, the percent yield is 84.2%.

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