Answer:
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%
Considering the reaction stoichiometry and the ideal gas law, the percent yield when 1.72 g of H₂O₂ decomposes and produces 375 mL of O₂ gas measured at 42 °C and 1.52 atm is 86.96%.
The balanced reaction is:
2 H₂O₂ → 2 H₂O + O₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The molar mass, this is the amount of mass a substance contains in one mole, of H₂O₂ is 34 [tex]\frac{g}{mole}[/tex]. Then, the amount of moles of H₂O₂ that decomposes when 1.72 grams of H₂O₂ reacts is calculated as:
[tex]1.72 gramsx\frac{1 mole}{34 grams}= 0.0506 moles[/tex]
Then you can apply the following rule of three: if by stoichiometry 2 moles of H₂O₂ produce 1 moles of O₂, 0.0506 moles of H₂O₂ will produce how many moles of O₂?
[tex]amount of moles of O_{2} =\frac{0.0506 moles of H_{2} O_{2} x1 mole of O_{2} }{2 moles of H_{2} O_{2}}[/tex]
amount of moles of O₂= 0.0253 moles
On the other side, an ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, for O₂ gas you know:
Replacing:
1.52 atm× 0.375 L = n× 0.082 [tex]\frac{atmL}{molK}[/tex]× 315 K
Solving:
[tex]n=\frac{1.52 atmx 0.375 L}{0.082\frac{atmL}{molK}x 315 K }[/tex]
n= 0.022 moles
The percent yield is calculated as
[tex]Percent yield= \frac{real yield}{theoretical yield} x100[/tex]
In this case, for oxygen the percent yield is calculated as
[tex]Percent yield of oxygen= \frac{0.022 moles}{0.0253 moles} x100[/tex]
Percent yield of oxygen= 86.96 %
Finally, the percent yield when 1.72 g of H₂O₂ decomposes and produces 375 mL of O₂ gas measured at 42 °C and 1.52 atm is 86.96%.
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