Question

Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide

Answer 1

Answer:

87.40 %

Explanation:

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

Step 1; moles of sodium oxide

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

          = 0.0807 moles

Step 2: Theoretical moles of sodium peroxide produced

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

Step 3: Theoretical mass of sodium peroxide used

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

                            = 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

Step 4: Percent yield of Na₂O₂

• We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

$Percent yield=(\frac{Actual yield}{theoretical yield})100$

$Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100$

                       = 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%