Question #f6afd

The idea here is that you need to use the stoichiometric coefficients that exist between the chemical species tha ttake part in the reaction to figure out how many molecules of reactants will actually take part in the reaction.

The balanced chemical equation for this reaction looks like this

#color(red)(2)"C"_4"H"_text(10(g]) + 13"O"_text(2(g]) -> color(blue)(8)"CO"_text(2(g]) + color(purple)(10)"H"_2"O"_text((g])#

Usually, you would look at the stoichiometric coefficients of the reactants and of the products as representing mole ratios. But since a mole is simply a very large collection of molecules, you can look at these ratios as being molecule ratios.

So, you have a #color(red)(2):13# molecule ratio between butane, #"C"_4"H"_10#, and oxygen gas. This tells you that in order for the reaction to take place, you need to have #13# molecules of oxygen gas for every #color(red)(2)# molecules of butane.

You know that you're starting with #30# molecules of butane and #30# molecules of oxygen gas. In order for all the molecules of butane to react, you would need

#30 color(red)(cancel(color(black)("molecules C"_4"H"_10))) * "13 molecules O"_2/(color(red)(2)color(red)(cancel(color(black)("molecules C"_4"H"_10)))) = "195 molecules of O"_2#

Since you have fewer molecules of oxygen gas, this compound will act as a limiting reagent, i.e. it will determine how many molecules of butane actually react.

More specifically, #30# molecules of oxygen would allow for

#30 color(red)(cancel(color(black)("molecules O"_2))) * (color(red)(2)" molecules C"_4"H"_10)/(13color(red)(cancel(color(black)("molecules O"_2)))) = "4.615 molecules C"_4"H"_10#

Now, it's important to realize that you're dealing with molecules, so you can't have decimal numbers, since that would imply fractions of a molecule.

So, you don't have enough oxygen to allow for #5# molecule of butane to react, so only #4# will actually react. This will consume

#4 color(red)(cancel(color(black)("molecules C"_4"H"_10))) * "13 molecules O"_2/(color(red)(2)color(red)(cancel(color(black)("molecules C"_4"H"_10)))) = "26 molecules of O"_2#

So, #26# molecules of oxygen and #4# molecules of butane will actually react. Pick one reactant and use the other molecule ratios to find how many molecules of products will result

#4 color(red)(cancel(color(black)("molecules C"_4"H"_10))) * (color(blue)(8)" molecules CO"_2)/(color(red)(2)color(red)(cancel(color(black)("molecules C"_4"H"_10)))) = "16 molecules CO"_2#

#4 color(red)(cancel(color(black)("molecules C"_4"H"_10))) * (color(purple)(10)" molecules H"_2"O")/(color(red)(2) color(red)(cancel(color(black)("molecules C"_4"H"_10)))) = "20 molecules H"_2"O"#

So, the reaction will consume

• #4# molecules of butane
• #26# molecules of oxygen gas

and produce

• #16# molecules of carbon dioxide
• #20# molecules of water

The molecules that have been consumed are no longer present in the container. Don't forget that we still have molecules of butane and of oxygen gas that did not take part in the reaction! More specifically, you will also have

• #30 -4 = 26# molecules of butane
• #30 - 26 = 4# molecules of oxygen gas

Therefore, the total number of molecules present in the reaction vessel after the reaction is completed will be

#overbrace(26 + 4)^("unreacted") + overbrace(16 + 20)^(color(brown)("produced")) = color(green)("66 molecules")#