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
