# Question #9d98c

##### 1 Answer

#### Explanation:

You can go about solving this problem in two ways

*by using the percent composition of calcium in calcium hypochlorite**by using mole ratios*

Let's take the first method. The percent composition of calcium in calcium hypochlorite,

#(40.078color(red)(cancel(color(black)("g/mol"))))/(142.98color(red)(cancel(color(black)("g/mol")))) xx 100 = 28.03%#

This means that *every 100 grams* of calcium hypochlorite will contain **28.03 grams** of calcium.

Now, use the concentration of the solution and its volume to figure out how many grams of calcium cations you have

#2color(red)(cancel(color(black)("L sol"))) * ("200 mg Ca"""^(2+))/(1color(red)(cancel(color(black)("L sol")))) = "400 mg Ca"""^(2+)#

This means that, in order to get this much calcium into the solution, you must have dissolved

#400 * 10^(-3)color(red)(cancel(color(black)("g Ca"))) * ("100 g Ca"("ClO")_2)/(29.03color(red)(cancel(color(black)("g Ca")))) = "1.427 g Ca"("ClO")_2#

You need to round this off to one sig fig, but I will leave the answer rounded to two sig figs.

#m = color(green)("1.4 g Ca"("ClO")_2)#

Let's double-check the result using the second method. You know that you have

#400 * 10^(-3)color(red)(cancel(color(black)("g Ca"^(2+)))) * "1 mole Ca"/(40.078color(red)(cancel(color(black)("g Ca"^(2+))))) = "0.00998 moles Ca"""^(2+)#

Now, *every mole* of calcium hypochlorite produces **1 mole** of calcium ions in solution

#"Ca"("ClO")_text(2(aq]) -> "Ca"_text((aq])^(2+) + 2"ClO"_text((aq])^(-)#

This means that the number of moles of calcium ions will be **euqal to** the number of moles of calcium hypochlorite that was dissolved to make the solution.

Use calcium hypochlorite's molar mass to find how many grams you needed

#0.00998 color(red)(cancel(color(black)("moles"))) * "142.98 g"/(1color(red)(cancel(color(black)("mole")))) = "1.427 g" = color(green)("1.4 g")#

The answer is confirmed.