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HIGHER GRADE CHEMISTRY CALCULATIONS Reacting Volumes. Since one mole of any gas occupies the same volume under the same conditions of temperature and pressure.

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Presentation on theme: "HIGHER GRADE CHEMISTRY CALCULATIONS Reacting Volumes. Since one mole of any gas occupies the same volume under the same conditions of temperature and pressure."— Presentation transcript:

1 HIGHER GRADE CHEMISTRY CALCULATIONS Reacting Volumes. 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 40cm 3 of nitrogen monoxide reacts with 100 cm 3 of oxygen? The balanced equation for the reaction is:- 2NO(g) + O 2 (g)  2NO 2 (g) 2 mol + 1 mol  2 mol 2 vol + 1 vol  2 vol 2 vol of NO = 40cm 3 So 1 vol = 20cm 3 40cm cm 3  40cm 3 20cm 3 of O 2 used So 80cm 3 left unreacted Volume composition = 40cm 3 of NO 2 (g) + 80cm 3 of unreacted O 2

2 Higher Grade Chemistry Calculations for you to try. 1.40cm 3 of propane is burned in 250cm 3 of oxygen. Calculate the volume and composition of the resulting gas mixture. (All measurements are made at room temperature and pressure). The balanced equation for the reaction is:- C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l) 1 mol + 5 mol  3 molThe water is ignored as at room temperature it is a liquid 1 vol + 5 vol  3 vol 40cm cm 3  120cm 3 As 200cm 3 of O 2 is used up there is 50cm 3 of O 2 left unreacted Volume composition = 120cm 3 of CO 2 (g) + 50cm 3 of unreacted O 2

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


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