daniaridha Posted September 28, 2014 Report Share Posted September 28, 2014 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. Reply Link to post Share on other sites More sharing options...
whatislife Posted September 29, 2014 Report Share Posted September 29, 2014 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 molesBack 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 6.5Hence, 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. Reply Link to post Share on other sites More sharing options...
Msj Chem Posted October 1, 2014 Report Share Posted October 1, 2014 A slightly easier way of thinking about it is to look at the ratio of C4H10 to O2, which is 2:13 or 1:6.5So 20 molecules of C4H10 needs (6.5x20) = 130 molecules of O2 to react completely. So O2 is the limiting reactant. Reply Link to post Share on other sites More sharing options...
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