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If you could please help explain the process of how to solve this question, that would be greatly appreciated

1. Butane lighters work by the release and combustion of pressurized butane.

2C4H10+13O2=8CO2+10H2O

Determine the limiting reagent in the following reaction.

a) 20 molecules of C4H10 and 100 molecules of O2.

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Well basically I think of it as working on ratios. To find the number of moles use the formula: no. of moles = number of molecules N divided by L (Avogadro's constant which is 6.02 x 10^23).
So, you've been given 3.32 x 10^-23 moles of butane.

Doing the same for oxygen you get 1.66 x 10^-22 moles

Back to the concept of ratios, you would need 2.16 x 10^-22 moles of oxygen in order to react with 3.32 x 10^-23 moles of butane. I did this by multiplying (3.32 x 10^-23) by 6.5

Hence, you don't have enough oxygen and so it's your limiting reagent!

I hope this is correct because if it's not, I'm screwed for finals lol.

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A slightly easier way of thinking about it is to look at the ratio of C4H10 to O2, which is 2:13 or 1:6.5

So 20 molecules of C4H10 needs (6.5x20) = 130 molecules of O2 to react completely.

So O2 is the limiting reactant.

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