Question

b. A 13.8 g sample H202 dissolved in water requires 52.5 mL of 0.095 M KMnO4 solution to reach the endpoint. What is the mass percent and concentration of H2O2 in the original sample? Assume that thę prügina males solution has a density of 1.00 g/mL

Answer 1

Number of moles of

Number of moles of� KMnO_4 = 52.5 mL/1000 mL/L times 0.095 mol/L Number of moles of KMnO_4 = 0.004988 mol 2KMnO_4 + 6H^+ + 5H_2O_2 rightarrow 2 Mn^2 + + 8 H_2O + 5O_2 + 2K^+ Number of moles of H_2O_2 = 5/2 times number of moles of KMnO_4 Number of moles of H_2O_2 = 5/2 times 0.004988 mol Number of moles of H_2O_2 = 0.0125 mol The molar mass of H_2O_2 = 34 g/mol Mass of H_2O_2 = 34 g/mol times 0.0125 mol Mass of H_2O_2 = 0.424 g Mass percent of H_2O_2 = 0.424 g times 100/13.8 g = 3.07 % Number of moles of H_2O_2 = 0.0125 mol Mass of sample is 13.8 g. its density is 1.00 g.mL. Hence, volume of sample is 13.8 g times 1.00 g/mL = 13.8 ml Concentration of H_2O_2 = 0.0125 mol/13.8 mL times 1000 mL/L Concentration of H_2O_2 = 0.904 M

Number of moles of

$2KMnO_4+6H^++5H_2O_2 \to 2Mn^{2+}+8H_2O+5O_2+2K^+$

Number of moles of $H_2O_2= \frac{5}{2} \times \text { number of moles of } KMnO_4$

Number of moles of $H_2O_2= \frac{5}{2} \times 0.004988 \ mol$

Number of moles of

The molar mass of

Mass of $H_2O_2= 34 \ g/mol \times 0.0125 \ mol$

Mass of

Mass percent of $H_2O_2= \frac{0.424 \ g \times 100 }{13.8 \ g}=3.07 \ \%$

Number of moles of

Mass of sample is 13.8 g. Its density is 1.00 g/mL. Hence, volume of sample is $13.8 \ g \times 1.00 \ g/mL = 13.8 \ ml$

Concentration of $H_2O_2= \frac{0.0125 \ mol}{13.8 \ mL} \times 1000 \ mL/L$

Concentration of $H_2O_2= 0.904 \ M \simeq 0.9 \ M$