Could someone help me solve this Chem problem? PLEASE!?

Question

Okay so the problem is: A 8.746 gram mixture of calcium hydroxide and potassium hydroxide is exposed to an atmosphere of pure CO2. The mass of solid increases to a final value of 14.472 grams. Calculate the percentages of the individual solids in the original mixture, given that the following reactions occur.

Ca(OH)2 (s) + CO2 (g) ==> CaCO3 (s) + H2O (g)
KOH (s) + CO2 (g) ==> KHCO3 (s)


Thanks so much. I dont really need you to show me every detail of how to do it, but just the start of a basic idea of what I'm supposed to do would do fine. thanks again!

Answer 1

molecular weights:
Ca(OH)2:74.093
KOH:56.106
CaCO3:100.087
KHCO3:100.11
let the number of moles of Ca(OH)2 and KOH in the initial mixture be x and y respectively. then the following equation follows:
74.093x + 56.106y = 8.746
assuming both reactions go to completion, the number of moles of CaCO3 and KHCO3 in the final mixture are also x and y respectively. Water leaves as vapor and as such is not contained in the 14.472 g.
therefore:
100.087x + 100.11y= 14.472

after that, it's all just a matter of solving the 2 equations for x and y and from that one can calculate mass precent in the initial mixture.

Answer 2

Ca(OH)2 ==> 40 + 32 + 2 = 74 g/m
CaCO3 ==> 40 + 12 + 48 = 100 g/m
KOH ==> 39 + 16 + 1 = 56 g/m
KHCO3 ==> 39 + 1 + 12 + 48 = 100 g/m
Let x = mass of Ca(OH)2
y = mass of KOH
x + y = 8.746
100x/74 + 100y/56 = 14.472
x/74 + y/56 = 0.14472
56x/74 + y = 8.10432
x - 56x/74 = 0.64168
74x - 56x = 47.48432
18x = 47.48432
x ≈ 2.6380 g ≈ 30.16% Ca(OH)2
y ≈ 6.1060 g ≈ 68.84% KOH

Answer 3

Let X = amount of Ca(OH); its weight will increase by the following faxtor { MW CaCO3 / MW Ca(OH)2}
Let 8.746 - X = weight of KOH; its weight will increase by the factor {MW KHCO3 / MW KOH}

X * (100.09/74.10) + (8.746 - X) * (100.1/56.1) = 14.472 g