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

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 Na₂O₂We 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%

At first, our organs can survive without oxygen and nutrients for a while. Only gradually does the cell division stop completely, then the cells die. If too many cells have died, the organs can no longer regenerate. The quickest reaction occurs in the brain, where the cells die after three to five minutes.

Hope it helps.