Question #b2d74
1 Answer
Explanation:
I'll show you how to solve this problem without using the formula given to you first, then double-check the result by using the formula.
Start by writing the balanced chemical equation that describes this combustion reaction
#"C"_ 3"H"_ (8(g)) + 5"O"_ (2(g)) -> color(blue)(3)"CO"_ (2(g)) + 4"H"_ 2"O"_ ((l))#
Notice that when
#1 color(red)(cancel(color(black)("mole C"_3"H"_8))) * "44.10 g"/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = "44.10 g"#
#3 color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "132.03 g"#
So, this tells you that for every
I say theoretically because this corresponds to a
Now, you are told that the reaction has a
Use this to find the mass of carbon dioxide that would correspond to a
#8.53 color(red)(cancel(color(black)("g CO"_2))) * ("100 g CO"_2color(white)(a)"theoretical")/(93color(red)(cancel(color(black)("g CO"_2color(white)(a)"actual")))) = "9.172 g CO"_2#
Now all you have to do is use the gram ratio that exists between propane and carbon dioxide to see what mass of propane would be needed to produce this much carbon dioxide
#9.172 color(red)(cancel(color(black)("g CO"_2))) * ("44.10 g C"_3"H"_8)/(132.03color(red)(cancel(color(black)("g CO"_2)))) = color(green)(bar(ul(|color(white)(a/a)color(black)("3.06 g C"_3"H"_8)color(white)(a/a)|)))#
I'll leave the answer rounded to three sig figs.
So, you have this formula
#color(purple)(bar(ul(|color(white)(a/a)color(black)("% yield" = "actual yield"/"theoretical yield" xx 100%)color(white)(a/a)|)))#
You know the percent yield and the actual yield, so your job here is to figure out what the theoretical yield would be.
Rearrange to find
#"theoretical yield" = "actual yield"/"% yield" xx 100%#
Plug in your values to find
#"theoretical yield" = "8.53 g"/(93color(red)(cancel(color(black)(%)))) xx 100color(red)(cancel(color(black)(%))) = "9.172 g"#
Once again, you know that when this reaction has a
At this point, you would use the gram ratio again to see how much propane would be needed to produce
As a final note, the general idea here is that the amount of propane needed to produce