If the reaction described by this chemical equation started with 5.77 grams of PbO2 and resulted in 0.331 grams of O2, what is the percent yield of O2

42.9%

Explanation:

Step 1: Write the balanced decomposition reaction

PbO₂ ⇒ Pb + O₂

Step 2: Calculate the theoretical yield of O₂ from 5.77 g of PbO₂

According to the balanced equation, the mass ratio of PbO₂ to O₂ is 239.2:32.00.

5.77 g PbO₂ × 32.00 g O₂/239.2 g PbO₂ = 0.772 g O₂

Step 3: Calculate the percent yield of O₂

The real yield of O₂ is 0.331 g. The percent yield of O₂ is:

%yield = real yield / theoretical yield × 100%

%yield = 0.331 g / 0.772 g × 100% = 42.9%

In the case of mixtures of ethanol and water, this minimum occurs with 95.6% by mass of ethanol in the mixture. The boiling point of this mixture is 78.2°C, compared with the boiling point of pure ethanol at 78.5°C, and water at 100°C. You might think that this 0.3°C doesn't matter much, but it has huge implications for the separation of ethanol / water mixtures. The next diagram shows the boiling point / composition curve for ethanol / water mixtures. I've also included on the same diagram a vapor composition curve in exactly the same way as we looked at on the previous pages about phase diagrams for ideal mixtures.