Mn or Co is treated with 1 mL of 6 M HNO3 followed immediately by the drop-wise addition of 3% hydrogen peroxide until the preci
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<h3>Answer:</h3>
87.40 %
<h3>Explanation:</h3>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
<h3>Step 1; moles of sodium oxide</h3>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
<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.
Thus, moles of sodium peroxide used is 0.0807 moles
<h3>Step 3: Theoretical mass of sodium peroxide used</h3>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
<h3>Step 4: Percent yield of Na₂O₂</h3>- We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.
= 87.40 %
Therefore, the percentage yield of sodium peroxide is 87.4%