In an experiment, calcium carbonate reacted with different volumes of hydrochloric acid in water. One of the products formed dur
ing the experiment was carbon dioxide. The time taken for 0.89 mL of carbon dioxide to form was recorded. A partial record of the experiment is shown. Experimental Record Flask Mass of Calcium Carbonate Volume of HCl Volume of Water Time 1 4.0 g 25 mL 0 mL 11.2 seconds 2 4.0 g 20 mL 5 mL 3 4.0 g 15 mL 10 mL 4 4.0 g 10 mL 15 mL
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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%