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%

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.