Hydrogen peroxide with a concentration of 3.0 percent (3.0 g of H2O2 in 100 mL of solution) is sold in drugstores for use as an antiseptic. For a 10.0-mL 3.0 percent H2O2 solution, calculate (a) the oxygen gas produced (in liters) at STP when the compound undergoes complete decomposition and (b) the ratio of the volume of O2 collected to the initial volume of the H2O2 solution.

Answer:

a) 0.099 L

b) 9.9

Explanation:

Now, given the equation for the decomposition of H2O2;

2H2O2(l) ------> 2H2O(l) + O2(g)

Mass of H2O2;

percent w/v concentration = mass/volume * 100

volume = 10.0-mL

percent w/v concentration = 3.0 percent

mass of  H2O2 = x

3 = x/ 10 * 100

30 = 100x

x = 30/100

x = 0.3 g of H2O2

Number of moles in  0.3 g of H2O2 = mass/ molar mass

Molar mass of H2O2 = 34.0147 g/mol

Number of moles in  0.3 g of H2O2 = 0.3g/34.0147 g/mol

= 0.0088 moles

From the reaction equation;

2 moles of H2O2 yields 1 mole of oxygen

0.0088 moles of H2O2 = 0.0088 * 1/2  = 0.0044 moles of oxygen

If 1 mole of oxygen occupies 22.4 L

0.0044 moles of oxygen occupies 0.0044 *  22.4/1

= 0.099 L

b) initial volume of the H2O2 solution = 10 * 10-3 L

Hence, ratio of the volume of O2 collected to the initial volume of the H2O2 solution = 0.099 L/10 * 10-3 L = 9.9

Which answer would represent 0.001 moles?