When 14.0 g of calcium metal is reacted with water, 5.00 g of calcium hydroxide is produced. Using the following balanced equation, calculate the percent yield for the reaction?
#"Ca"(s) + 2"H"_2"O"(l) -> "Ca"("OH")_2(aq) + "H"_2(g)#
I'm more so worried about the execution rather than the answer so if you do not include the answer it is fine! Thanks so much :D
I'm more so worried about the execution rather than the answer so if you do not include the answer it is fine! Thanks so much :D
1 Answer
Explanation:
In order to find the percent yield of the reaction, you need to know two things
- how many grams of calcium hydroxide would theoretically be produced by the reaction of
#"14.0 g"# of calcium metal with excess water#-># this is the theoretical yield of the reaction- how many grams of calcium hydroxide are actually produced by this reaction
#-># this is the actual yield of the reaction
The percent yield is calculated using the equation
#"% yield" = "actual yield"/"theoretical yield" * 100%#
Now, you know that the reaction produced
#"actual yield = 5.00 g"#
To find the theoretical yield of the reaction, use the fact that the reaction produces
#overbrace("Ca"_ ((s)))^(color(blue)("1 mole reacts")) + 2"H"_ 2"O"_ ((l)) -> overbrace("Ca"("OH")_ (2(aq)))^(color(blue)("1 mole is produced")) + "H"_ (2(g))#
Convert the mass of calcium to moles by using the molar mass of the metal.
#14.0 color(red)(cancel(color(black)("g"))) * "1 mole Ca"/(40.078color(red)(cancel(color(black)("g")))) = "0.3493 moles Ca"#
Since you're looking for the theoretical yield of the reaction, i.e. what you would expect to get at
#0.3493 color(red)(cancel(color(black)("moles Ca"))) * ("1 mole Ca"("OH")_2)/(1color(red)(cancel(color(black)("mole Ca")))) = "0.3493 moles Ca"("OH")_2#
Convert the number of moles of calcium hydroxide to grams by using the molar mass of the compound.
#0.3493 color(red)(cancel(color(black)("moles Ca"("OH")_2))) * "74.093 g"/(1color(red)(cancel(color(black)("mole Ca"("OH")_2)))) = "25.88 g"#
This means that at
Since the reaction produced
#"% yield" = (5.00 color(red)(cancel(color(black)("g"))))/(25.88color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(19.3%)))#
The answer is rounded to three sig figs.