What is the percent yield when 1.72 g of H2O2 decomposes and produces 375 mL of O2 gas measured at 42 oC and 1.52 atm

87.0%

Explanation:

Step 1: Write the balanced reaction

H₂O₂ ⇒ H₂O + 0.5 O₂

Step 2: Calculate the real yield of oxygen, in grams

We have 375 mL (0.375 L) of O₂ at 42 °C (315 K) and 1.52 atm. First, we will calculate the number of moles using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.52 atm × 0.375 L / (0.0821 atm.L/mol.K) × 315 K = 0.0220 mol

The molar mass of oxygen is 32.00 g/mol.

0.0220 mol × 32.00 g/mol = 0.704 g

Step 3: Calculate the theoretical yield of oxygen, in grams

According to the balanced equation, the mass ratio of H₂O₂ to O₂ is 34.01:16.00.

1.72 g H₂O₂ × 16.00 g O₂/34.01 g H₂O₂ = 0.809 g O₂

Step 4: Calculate the percent yield of oxygen

We will use the following expression.

%yield = real yield / theoretical yield × 100%

%yield = 0.704 g / 0.809 g × 100% = 87.0%