Answer:
1. Theoretical yield of NaOH is 22.72 g
2. Percentage yield of NaOH = 22.14%
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
NaHCO₃ —> NaOH + CO₂
From the balanced equation above,,
1 mole of NaHCO₃ decomposed to produce 1 mole (i.e 40 g) of NaOH and 1 mole (i.e 44.01 g) of CO₂.
Next, we shall determine the number of mole of NaHCO₃ that will decompose to produce 25 g of CO₂. This can be obtained as follow:
From the balanced equation above,,
1 mole of NaHCO₃ decomposed to produce 44.01 g of CO₂.
Therefore, Xmol of NaHCO₃ will decompose to 25 g of CO₂ i.e
Xmol of NaHCO₃ = 25 / 44.01
Xmol of NaHCO₃ = 0.568 mole
1. Determination of the theoretical yield of NaOH.
From the balanced equation above,,
1 mole of NaHCO₃ decomposed to produce 40 g of NaOH.
Therefore, 0.568 mole of NaHCO₃ will decompose to produce = 0.568 × 40 = 22.72 g of NaOH.
Thus, the theoretical yield of NaOH is 22.72 g
2. Determination of the percentage yield of NaOH.
Theoretical yield of NaOH = 22.72 g
Actual yield of NaOH = 5.03 g
Percentage yield of NaOH =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield = 5.03 / 22.72 × 100
Percentage yield of NaOH = 22.14%
= 
= 9.95 mL
x 100 
x 100 = 0.5 % error