Question

Ozzie wanted to do another experiment with a stronger H2O2 solution to check the accuracy of the experiment by calculating the theoretical volume of O2(g) it would produce. Then he could compare his experimental volume of O2(g) to the theoretical volume of O2(g). He used 5.20 mL of 3.18 M H2O2 and the partial pressure of O2 was 0.9959 atm and the temperature was 291.25 K. What volume of O2(g) could he theoretically produce (in mL)? Enter your answer to 3 significant figures.

Answer 1

Using the ideal gas law, Ozzie can theoretically produce approximately 2.16 mL of O2(g) from 5.20 mL of 3.18 M H2O2 solution at 291.25 K and 0.9959 atm partial pressure.

Step-by-step explanation:

Ozzie conducted an experiment to determine the theoretical volume of O2(g) produced from a 3.18 M H2O2 solution. To calculate this, we can use the ideal gas law, PV = nRT, where P is the partial pressure of O2, V is the volume we want to find, n is the number of moles of O2, R is the ideal gas constant, and T is the temperature in kelvin.

First, we calculate the number of moles of H2O2:

• n(H2O2) = volume (in liters) × molarity
• n(H2O2) = 0.00520 L × 3.18 M = 0.016536 moles

According to the balanced reaction 2H2O2(l) → 2H2O(l) + O2(g), 2 moles of H2O2 produce 1 mole of O2. So:

• n(O2) = ½ × n(H2O2)
• n(O2) = 0.008268 moles

Now we can use the ideal gas law:

• V(O2) = (n(O2) × R × T) / P
• V(O2) = (0.008268 moles × 0.0821 L·atm/mol·K × 291.25 K) / 0.9959 atm
• V(O2) = 2.16 mL (to three significant figures)

Ozzie can expect to produce a theoretical volume of approximately 2.16 mL of O2(g) under the given conditions. This value will help in evaluating the accuracy of the experimental results.