Question #9d98c

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, #"Ca"("ClO")_2#, is the ratio between the molar mass of calcium and the molar mass of the compound, multiplied by 100.

#(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 #"300 mg"# of #"Ca"^(2+)# cations in solution. Use calcium's molar mass to find the number of moles of calcium cations

#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.