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%.

The gas would begin to expand, and eventually the can would burst or explode.

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