HIGHER GRADE CHEMISTRY CALCULATIONS

1. The balanced equation for the reaction is:- 2NO(g) + O2(g)  2NO2 (g) HIGHER GRADE CHEMISTRY CALCULATIONS Since one mole of any gas occupies the same volume under the same conditions of temperature and pressure we can use the balanced equation to calculate the volume of gases. Worked example 1. Nitrogen monoxide reacts with oxygen to form nitrogen dioxide. What is the volume composition of the gases present when 40cm3 of nitrogen monoxide reacts with 100 cm3 of oxygen? Reacting Volumes. 2 mol + 1 mol  2 mol 2 vol of NO = 40cm3 So 1 vol = 20cm3 2 vol + 1 vol  2 vol 40cm3 + 20cm3 40cm3 20cm3 of O2 used So 80cm3 left unreacted Volume composition = 40cm3 of NO2 (g) + 80cm3of unreacted O2

2. The balanced equation for the reaction is:- C3H8(g) + 5O2(g)  3CO2 (g) + 4H2O(l) • Calculations for you to try. • 40cm3 of propane is burned in 250cm3 of oxygen. • Calculate the volume and composition of the resulting gas mixture. • (All measurements are made at room temperature and pressure). 1 mol + 5 mol  3 mol The water is ignored as at room temperature it is a liquid 1 vol + 5 vol  3 vol As 200cm3 of O2 is used up there is 50cm3 of O2 left unreacted 40cm3 + 200cm3 120cm3 Volume composition = 120cm3 of CO2 (g) + 50cm3of unreacted O2 Higher Grade Chemistry

3. The balanced equation for the reaction is:- CO(g) + 2H2(g)  CH3OH(g) Calculations for you to try. 2. 200 litres of carbon monoxide is reacted with 500 litres of hydrogen to form gaseous methanol, CH3OH. Calculate the volume and composition of the resulting gas mixture. (All measurements carried out at 200oC) 1 mol + 2 mol  1 mol 1 vol + 2 vol  1 vol 200 litres + 400 litres 200 litres As 400 litres of H2 is used up there is 100 litres of H2 left unreacted Volume composition = 200 litres of CH3OH(g) + 100 litres of H2 left unreacted Higher Grade Chemistry