Question #e8964

Start by taking a look at the balanced chemical equation given to you

#"C"_ 3"H"_ (8(g)) + color(blue)(4)"Cl"_ (2(g)) -> "C"_ 3"H"_ 4 "Cl"_ (4(g)) + color(purple)(4)"HCl"_ ((g))#

Notice that every mole of propane that takes part in the reaction requires #color(blue)(4)# moles of chlorine gas and produces #color(purple)(4)# moles of hydrogen chloride.

In your case, #4.0# moles of propane are said to react completely with #24# moles of chlorine gas. The problem mentions that one of the reactants has been completely consumed by the reaction.

As you can see, you have more moles of chlorine gas than you would need, since #4.0# moles of propane would only require

#4.0 color(red)(cancel(color(black)("moles C"_3"H"_8))) * (color(blue)(4)color(white)(.)"moles Cl"_2)/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = "16 moles Cl"_2#

This implies that propane acts as a limiting reagent, i.e. it is completely consumed before all the moles of chlorine gas get the chance to react.

So, you know for a fact that all the moles of propane react. You can now use the #1:color(purple)(4)# mole ratio that exists between propane and hydrogen chloride to find the number of moles of the latter produced by the reaction

#4.0 color(red)(cancel(color(black)("moles C"_3"H"_8))) * (color(purple)(4)color(white)(.)"moles HCl")/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = color(darkgreen)(ul(color(black)("16 moles HCl")))#

The answer is rounded to two sig figs.