# 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?

## ${\text{Ca"(s) + 2"H"_2"O"(l) -> "Ca"("OH")_2(aq) + "H}}_{2} \left(g\right)$ 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

Feb 10, 2018

19.3%

#### 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 $\text{14.0 g}$ of calcium metal with excess water $\to$ this is the theoretical yield of the reaction
• how many grams of calcium hydroxide are actually produced by this reaction $\to$ 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 $\text{5.00 g}$ of calcium hydroxide, so you can say that

$\text{actual yield = 5.00 g}$

To find the theoretical yield of the reaction, use the fact that the reaction produces $1$ mole of calcium hydroxide for every $1$ mole of calcium metal that reacts.

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 100% yield, you can use the $1 : 1$ mole ratio that exists between calcium and calcium hydroxide (no information was provided about the mass of water that takes part in the reaction, so you can presume that water is in excess) to say that the reaction would theoretically produce

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 100% yield, the complete reaction of $\text{14.0 g}$ of calcium with enough water would produce $\text{25.88 g}$ of calcium hydroxide.

Since the reaction produced $\text{5.00 g}$, you can say that the percent yield was equal to

"% 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.