Question

Dilute aqueous hydrogen peroxide is used as a bleaching agent and for disinfecting surfaces and small cuts. its concentration is sometimes given as a certain number of "volumes hydrogen peroxide," which refers to the number of volumes of o2 gas, measured at stp, that a given volume of hydrogen peroxide solution will release when it decomposes to o2 and liquid h2o. how many grams of hydrogen peroxide are in 0.125−l of 20 "volumes hydrogen peroxide" solution

Answer 1

7.62 grams.

Explanation

According to the question, 0.125 L of the 20 "volumes hydrogen peroxide" solution would decompose to produce $0.125 \times 20 = 2.5 \; \text{L}$ of oxygen gas.

The question indicates that the volume of $\text{O}_2$ here is measured under STP, which means that every $\text{mol}$ of gas would have a volume of $22.4 \; \text{L}$. There would thus be $2.5 / 22.4 = 0.112 \; \text{mol}$ of molecules in the 2.5 L of $\text{O}_2$ produced.

$2\; \text{H}_2\text{O}_2\; (aq)\to 2\; \text{H}_2\text{O} \; (l) + \text{O}_2 \; (g)$

It takes two moles of hydrogen peroxide to produce one mole of oxygen. The unknown solution must contain 0.224 mol of hydrogen peroxide to produce all the 0.112 mol of oxygen.

Hydrogen peroxide has a formula mass of $34.02$, meaning that each mole of it would have a mass of 34.02 grams. 0.224 mol of $\text{H}_2\text{O}_2$ would thus have a mass of $0.224 \times 34.02 = 7.62 \; \text{g}$

Answer 2

Answer:

7.56 g of hydrogen peroxide present in 0.125 L

Explanation:

The reaction is:

2H2O2 = O2 + 2H2O

Using the gas ideal to obtain the number of moles of O2 present in 1 liter:

$n=\frac{PV}{RT} =\frac{1*20}{0.08206*273} =0.89 moles$

The amount of hydrogen peroxide is:

$mH_{2}O_{2} =0.89molO_{2} *\frac{2molH_{2}O_{2} }{1molO_{2} } *\frac{34.01gH_{2}O_{2}}{1molH_{2}O_{2}} =60.5 g$

Mass of hydrogen peroxide present in 0.125L

$mH_{2}O_{2}=60.5\frac{g}{L} *0.125L=7.56 g$