# 24 g of carbon makes 14 g of carbon monoxide. What is the predicted yield and the percentage yield?

Jun 13, 2016

$\text{56 g CO}$

"% yield" = 25%

#### Explanation:

Start by writing the balanced chemical equation that describes the reaction between carbon and oxygen gas to produce carbon monoxide, $\text{CO}$

$\textcolor{red}{2} {\text{C"_ ((s)) + "O"_ (2(g)) -> 2"CO}}_{\left(g\right)}$

Notice that the reaction consumes $\textcolor{red}{2}$ moles of carbon and produces $2$ moles of carbon monoxide, so you can say that for a 100% yield reaction, the number of moles of carbon monoxide produced will be equal to the number of moles of carbon consumed.

The problem provides you with grams of carbon and of carbon monoxide, so use their respective molar masses to convert the grams to moles

24 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "2.0 moles C"

14 color(red)(cancel(color(black)("g"))) * "1 mole CO"/(28.01color(red)(cancel(color(black)("g")))) = "0.50 moles CO"

So, the theoretical yield of the reaction corresponds to what you'd get for 100% yield. If all the moles of carbon take part in the reaction, you should expect to get

2.0 color(red)(cancel(color(black)("moles C"))) * "2 moles CO"/(color(red)(2)color(red)(cancel(color(black)("moles C")))) = "2.0 moles CO"

Expressed in grams, the theoretical yield is

2.0 color(red)(cancel(color(black)("moles CO"))) * "28 g"/(1color(red)(cancel(color(black)("mole CO")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("56 g CO")color(white)(a/a)|)))

However, you know that the reaction produced only $0.50$ moles of carbon monoxide. This is the reaction's actual yield, which is simply what is actually produced by the reaction.

Percent yield is defined as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% yield" = "what you actually get"/"what you expect to get} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The percent yield of the reaction will be

"% yield" = (0.50 color(red)(cancel(color(black)("moles"))))/(2.0color(red)(cancel(color(black)("moles")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)color(black)("25%")color(white)(a/a)|)))