50 dm^3 of a mixture containing 60% CO and 40% H2, by volume, ifs completely burnt in oxygen.

Assuming that all volumes are measured at the same temperature and pressure,

a) what is the volume of the Oxygen required, abs

b) what is the volume of CO2 and steam, taken together, produced?

he vol measure used in hydrogen peroxide refers to the volume of oxygen that it produces per volume of hydrogen peroxide used up at room temperature.

So, if you consider the equation:

2H2O2 - -> 2H2O + O2

You can see that 2 moles of hydrogen peroxide produces 1 mole of oxygen, i. e. the number of moles of oxygen produced is half that of hydrogen peroxide used up.

If your hydrogen peroxide solution is 100 vol (quite reasonable for a strong stock solution) this means that 1000 cm3 of the H2O2 produces 100,000 cm3 of oxygen.

100 dm3 of oxygen = 100/24 moles = 4.17 moles

If 4.17 moles of oxygen are produced then there must have been 4.17 x 2 moles of hydrogen peroxide = 8.34 moles in 1 dm3 = 8.34 molar.

so if 100 vol is 8.34 molar then 1 vol = 0.0834 molar etc