Question

Question #b2d74

Answer 1

`"3.06 g"`

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` mole of propane reacts, `color(blue)(3)` moles of carbon dioxide are produced. To convert this to grams, use the molar masses of propane and carbon dioxide

`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 `"44.10 g"` of propane that undergo combustion, the reaction theoretically produces `"132.03 g"` of carbon dioxide.

I say theoretically because this corresponds to a `100%` yield, i.e. if all the grams of propane react and actually produce the maximum number of grams of carbon dioxide.

Now, you are told that the reaction has a `93%` yield, which means that for every `"100 g"` of carbon dioxide that could be theoretically produced, the reaction actually produces `"93 g"`.

Use this to find the mass of carbon dioxide that would correspond to a `100%` yield

`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 `93%` yield and it produces `"8.53 g"` of carbon dioxide, it could theoretically produce `"9.172 g"` of carbon dioxide.

At this point, you would use the gram ratio again to see how much propane would be needed to produce `"9.172 g"` of carbon dioxide at `100%` yield.

As a final note, the general idea here is that the amount of propane needed to produce `"9.172 g"` of carbon dioxide at `100%` yield will only produce `"8.53 g"` of carbon dioxide at `93%` yield.