Carbon monoxide combines with oxygen to make co2. You start with 2 moles of each reactant.a.) how many liters of co2 are made b.) one reactant is not used up. Which one? How many liters left over?...
Carbon monoxide combines with oxygen to make co2. You start with 2 moles of each reactant.
a.) how many liters of co2 are made
b.) one reactant is not used up. Which one? How many liters left over?
thank you so much for answering! :)
This all starts with the chemical equation:
CO + O2 -> CO2
To balance the equation, we need to make sure that there are the same number of Carbon and Oxygen atoms on both sides, which we don't (Carbons match, but there are 3 Oxygens in the reactants' side, and 2 Oxygens in the products' side).
Recognize that if you only have half of an O2 molecule, it becomes balanced:
CO + 1/2 O2 -> CO2
However, lots of chemistry teachers flip out if you have a fraction in a chemical equation, so we'll just multiply each term by 2 to make everything a nice integer:
2 CO + O2 -> 2 CO2
Just recheck to make sure all the amounts of atoms match up:
Reactants: Carbons = 2, Oxygens = 4
Products: Carbons = 2, Oxygens = 4
Now, we can use this equation to find out how much CO2 is made.
Notice that for every O2 molecule, 2 CO molecules are used up. So this means even though you have 2 moles of each gas, you'll be using 2 moles of CO to react with 1 mole of O2. Even though you have 2 moles of O2 to react, only one mole can take part in the reaction. (This will be important for the second part)
Now, also notice from this equation, for every 2 moles of CO reacted, you can produce 2 moles of CO2. Because you only have 2 moles of CO, the most CO2 you can produce is 2 moles.
So, let's just say you make the most CO2 you possibly can: 2 moles. Now, we're not given much information about temperature and pressure, so let's just make an assumption that we're at STP: 0 degrees C, 1 atm. At STP, 1 mole of any ideal gas takes up 22.4 L. Because we have 2 moles of CO2, we can easily find the volume with the following equation:
Volume(CO2) = Moles(CO2) * 22.4L (at STP) = 2 * 22.4
Volume(CO2) = 44.8 L
Now, remember how our equation was
2 CO + O2 -> 2 CO2
So, for every mole of O2, 2 moles of CO are reacted. Which means even though we have 2 moles of O2, we can only use 1 mole. The CO is all used up by the time we get to the second mole!
For this reason, O2 is the reactant "in excess" (the one we have too much of). To find the volume, we use the same relationship we used to find the liters of CO2, except now insteaed of using moles of product, we're using leftover moles of reactant:
Volume(O2) = Leftover Moles(O2) * 22.4 L = 1 * 22.4 L
Volume(O2) = 22.4 L
Of course, if the problem gives you other conditions, I'll be glad to edit this post to reflect the different pressure and temperature conditions. It might be good to read about STP, anyway, though. It's used all the time in chemistry, and we're nowhere near summer!