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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

first we get the MM of each chemical species
CaCl2 = 110.9 g/mol
KOH = 56.11 g/mol
Ca(OH)2 = 74.016 g/mol
then get the limiting reactant by given mass of the reactants by each molecular mass considering the stoichiometric ratio
0.392 g CaCl2 (1mol /110.9 g)CaCl2*(1moleCa(OH)2/1mole CaCl2)
= 3.535 x 10^-3 mol Ca(OH)2
0.515 g KOH (1mol /56.11 g)KOH*(1moleCa(OH)2/2mole KOH)
= 4.589 x 10^-3 mol Ca(OH)2
thus our limiting reactant was CaCl2
using the calculated mole of Ca(OH)2 from CaCl2 we get the mass of the Ca(OH)2
mass = 3.535 x 10^-3 mole Ca(OH)2 * (74.016g/1mole )Ca(OH)2
= 0.262 g Ca(OH)2

best answer?

Answer 2

This is a very common kind of chemistry problem dealing with "Limiting Reactants".

You first need to calculate the number of moles of each of the two starting compounds that you have. From each of their masses, using their molar masses, you can calculate moles. Now, from the balanced equation you see that 1 mole of CaCl2 can yield one mole of Ca(OH)2, and that you need 2 moles of KOH to give one mole of Ca(OH)2. So, you can determine which of the two starting products will give you the smallest number of moles of the desired product. THAT one is your limiting reactant, and you cannot make more moles of the product than the limiting reactant will allow. Finally, you can use the molar mass of Ca(OH)2 to convert that number of moles of Ca(OH)2 into a mass.

That should get you through this kind of problem in the future...

Answer 3

CaCI2+45