Can you help me? i really don't understand this problem?

Question

CaCl2(aq) + 2 KOH(aq)Ca(OH)2(s) + 2 KCl(aq)

CaCl2 is provided in the form CaCl2·2H2O, but upon dissolving it in water, the waters of hydration can be ignored. There is a 1:1 stoichiometric relationship between CaCl2·2H2O and CaCl2.

0.392 grams of CaCl2·2H2O are combined with 0.515 grams of KOH. What is the theoretical yield of Ca(OH)2(s) in grams?

Answer 1

CaCl2(aq) + 2 KOH(aq) -> Ca(OH)2(s) + 2 KCl(aq)

So, per 2 mol KOH put in, or 1 mol CaCl2, you get 1 mol Ca(OH)2 out.
1 mol KOH = 19 + 8 + 2 = 29g (2 mol = 58g)
1 mol CaCl2·2H2O = 20 + 2*17 + 4*2 + 2* 8 = 74g

.392g CaCl2·2H2O / 74g/mol = 0.005297 mol
.515h KOH / 29g/mol = 0.01776 mol

Since there is 2 mol of KOH used per mol of CaCl2, and there the molar amount of CaCl2 is less than half of the molar amount of KOH, then CaCl2 is your limiting factor in the reaction.

If there is 0.005297 mol of CaCl2, per the reaction, you get 0.005297 mol of Ca(OH)2 as a product.

Ca(OH)2 = 20 + 18 = 38g/mol
0.005297 mol * 38g/mol = 0.201g Ca(OH)2 in product (solution!)

Answer 2

Answer: KOH is in excess. Theoretical yield is 0.197g of Ca(OH)2

Answer 3

first write the net ionic reaction:

Ca2+ + 2Cl- + 2K+ 2OH- --> Ca(OH)2 + 2K+ + 2Cl-

Ca2+ + 2OH- --> Ca(OH)2

Find the mass percentage of Ca(OH)2 that can be produced from Ca2+ and 2OH-.

first find moles of CaCl2.2H2O (grams * molecular weight)
then find moles of CaCl2 (stoich relationship of 1:1) and moles of KOH
find the limiting reagent (which reactant gets consumed completely).
result will be moles of Ca(OH)2 produced.
use mass percentage of: mass Ca(OH)2 divided by total mass of reactants times 100%