2 Na202 + 2H20 -> 4NaOH + 02 When 200.g of Na202 were reacted, 32.0 g of 02 were collected. The yield of 02 collected was (how many percentages)
To find the yield of O2 collected in percentage, you need to calculate the actual yield of O2 and divide it by the theoretical yield, then multiply by 100.
First, determine the molar mass of Na2O2:
Na2O2 = (2 x atomic mass of Na) + atomic mass of O
= (2 x 22.99 g/mol) + 16.00 g/mol
= 45.98 g/mol + 16.00 g/mol
= 61.98 g/mol
Next, calculate the moles of Na2O2 reacted:
moles of Na2O2 = mass of Na2O2 / molar mass of Na2O2
= 200 g / 61.98 g/mol
≈ 3.226 mol
From the balanced chemical equation, you can see that the stoichiometric ratio between Na2O2 and O2 is 2:1. Therefore, the theoretical yield of O2 is half the moles of Na2O2 reacted:
theoretical yield of O2 = 0.5 x moles of Na2O2
= 0.5 x 3.226 mol
≈ 1.613 mol
Now, calculate the actual yield of O2 collected, which is given as 32.0 g.
Next, calculate the yield percentage:
yield percentage = (actual yield / theoretical yield) x 100
= (32.0 g / (1.613 mol x 32.00 g/mol)) x 100
≈ 100%
Therefore, the yield of O2 collected is approximately 100%.