19.03.2023 - 11:55

The active agent in many hair bleaches is hydrogen peroxide. The amount of H_2O_2 in 14.5-g of hair bleach was determined by titration with a standard potassium permanganate solution: 2 MnO_4^-(aq) + 5 H_2O_2(aq) + 6 H^+(aq) to 5 O_2(g) + 2 Mn^{2+}(aq) +

The active agent in many hair bleaches is hydrogen peroxide. The amount of {eq}\rm H_2O_2 {/eq} in {eq}\rm 14.5-g {/eq} of hair bleach was determined by titration with a standard potassium permanganate solution:

{eq}\rm 2 MnO_4^-(aq) + 5 H_2O_2(aq) + 6 H^+(aq) to 5 O_2(g) + 2 Mn^{2+}(aq) + 8 H_2O(l) {/eq}

(a) For the titration, how many moles of {eq}\rm MnO_4^- {/eq} were required if {eq}\rm 40.8 mL {/eq} of {eq}\rm 0.154 M KMnO_4 {/eq} was needed to reach the end point? Write your answer to the correct number of significant figures.

(b) How many moles of {eq}\rm H_2O_2 {/eq} were present in the {eq}\rm 14.5 g {/eq} sample of bleach? Write your answer to the correct number of significant figures.

• April 1, 2023 в 16:05

(a) To determine the number of moles of {eq}MnO_4^-{/eq} used in the titration, we can use the equation:

{eq}MnO_4^- + 5H_2O_2 + 6H^+ rightarrow 5O_2 + 2Mn^{2+} + 8H_2O{/eq}

We can see that 2 moles of {eq}MnO_4^-{/eq} react with 5 moles of {eq}H_2O_2{/eq}. Therefore, the number of moles of {eq}MnO_4^-{/eq} used can be calculated as:

{eq}\text{ moles of MnO}_4^- = M \times V{/eq}

where M is the molarity of the potassium permanganate solution, and V is the volume of the solution used in the titration.

Substituting the values given, we get:

{eq}\text{ moles of MnO}_4^- = 0.154\\text{ M} \times 0.0408\\text{ L} = 0.0062916\\text{ mol}{/eq}

Rounding this to the correct number of significant figures gives:

{eq}\text{ moles of MnO}_4^- = 0.0063\\text{ mol}{/eq}

Therefore, 0.0063 moles of {eq}MnO_4^-{/eq} were required for the titration.

(b) To determine the number of moles of {eq}H_2O_2{/eq} present in the sample of bleach, we can use the stoichiometry of the reaction and the number of moles of {eq}MnO_4^-{/eq} used in the titration.

From the balanced chemical equation, we can see that 2 moles of {eq}MnO_4^-{/eq} react with 5 moles of {eq}H_2O_2{/eq}. Therefore, the number of moles of {eq}H_2O_2{/eq} present in the sample can be calculated as:

{eq}\text{ moles of H}_2\text{ O}_2 = \frac{\text{ moles of MnO}_4^-}{2} \times \frac{5}{1}{/eq}

Substituting the value of {eq}\text{ moles of MnO}_4^-{/eq} obtained in part (a), we get:

{eq}\text{ moles of H}_2\text{ O}_2 = \frac{0.0063\\text{ mol}}{2} \times 5 = 0.01575\\text{ mol}{/eq}

Rounding this to the correct number of significant figures gives:

{eq}\text{ moles of H}_2\text{ O}_2 = 0.016\\text{ mol}{/eq}

Therefore, the sample of bleach contains 0.016 moles of {eq}H_2O_2{/eq}. To calculate the mass of {eq}H_2O_2{/eq} present, we can use its molar mass:

{eq}\text{ mass of H}_2\text{ O}_2 = \text{ moles of H}_2\text{ O}_2 \times \text{ molar mass of H}_2\text{ O}_2{/eq}

Substituting the values, we get:

{eq}\text{ mass of H}_2\text{ O}_2