When CO (g) (0.3013 mol/L) and 183.2 grams of O 2 (g) in a 38.00 L reaction vess

Question

When CO(g)(0.3013 mol/L) and 183.2 grams of O2(g)in a 38.00 L reaction vessel at 532.0 When CO(g)(0.3013 mol/L) and 183.2 grams of O2(g)in a 38.00 L reaction vessel at 532.0

Explanation / Answer

ok well standard temperature and pressure is stp 22.4 litres per mole of gas. first write a balanced chemical equation 2C + 02 ------> 2CO then write mole ratios 2C + 02 ------> 2CO 2 : 1 : 2 using gay lussacs law and avogadros hypothesis we know that the same volumes of gases in same conditions are mole ratios (only works with gases however) so therefore the volume of CO gas produced will be 2 x that of the volume of oxygen gas (because ratio of O2 gas to CO gas is 1:2) =2x2.24 =4.48 litres of CO gas produced then the moles can be worked out by - number of moles = volume given/volume of 1 mole at STP = 4.48/22.4 =0.2 moles roughly i think. Answer: 1.20 g C is 1/10 mole of C 2.24 liters of O2 is 1/10 mole of O2 So when equal volumes molar anounts of C and O2 react: C (1 mole)+ O2(1 mole)-> CO2 (1 mole) (No CO is produced) With the given initial amounts 1/10 mole (2.24 liters) of CO2 would be produced If oxygen were limited all C could be converted to CO. This would rturn the inital amount of C (1,2 g) and 1.12 liters of O2 into 1/10 mole of CO (2.24 liters)

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