In the following combustion reaction of acetylene (C2H2), how many liters of CO2 will be produced if 60 liters of O2 is used, given that both gases are at STP?
The volume of one mole of gas at STP is 22.4 liters.
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First, and always, write a balanced equation for the reaction:
2 C2H2 + 5 O2 ----------> 4 CO2 + 2 H2O
I'm assuming that when they say that 60 liters of oxygen are used they mean that 60 liters of oxygen all reacted with the acetylene. If the gasses are all at the same temperature and pressure the problem becomes pretty simple. You're supposed to know, Dany, and I hope you do, that equal volumes of gas at the same temperature and pressure contain equal numbers of molecules. So, looking at the balanced equation, for every 5 liters of oxygen you will produce 4 liters of CO2. Since you have 60 liters of oxygen and the ratio of oxygen to CO2 is 5 to 4 you will get 4/5 times as much CO2 as you have O2 or 4/5 x 60 = 48 liters
Here's the thing: 5 O2 -----> 4 CO2
5 / 5 O2 ------> 4 / 5 CO2 .....or for every one volume (or mole) of O2 you get 4 / 5 ths volune (or mole) of CO2.
Hopefully, this helps you