24 g of carbon makes 14 g of carbon monoxide. What is the predicted yield and the percentage yield?
1 Answer
Explanation:
Start by writing the balanced chemical equation that describes the reaction between carbon and oxygen gas to produce carbon monoxide,
#color(red)(2)"C"_ ((s)) + "O"_ (2(g)) -> 2"CO"_ ((g))#
Notice that the reaction consumes
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
#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
Percent yield is defined as
#color(blue)(|bar(ul(color(white)(a/a)"% yield" = "what you actually get"/"what you expect to get" xx 100 color(white)(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)|)))#