The first step is to write a balanced chemical equation.
[tex]Ca(OH)_2+\text{ 2HCl}\rightarrow CaCl_2+2H_2O[/tex]The above equation is already balanced.
Now lets write what we have.
Ca(OH)2 = 1.95g
CaCl2 = 2.50g
We already know that HCl is in excess, meaning Ca(OH)2 is the limiting reagent.
%Yield = (actual yield/theoretical yield) x 100
Lets first calculate the theoretcal yield. To find this we need to first calculate the number of moles of CaCl2.
n = m/M where m is the mass and M is the molar mass of CaCl2
n = 2.50g/110,98 g/mol
n = 0.0225 mol
Theoretical Yield is the amount of product that would have been produced if all of the limiting reagent reacted and it was 100% pure. The limiting reagent is used up first in a reaction and controls the amount of product that can be produced.
The actual yield of CaCl2 we were given, which is 2.50g of CaCl2.
For every 1 mole of Ca(OH)2, 1 mole of CaCl2 is produced
n = 1.95g/74.09 g/mol
n = 0.0263 mol
Therefore the number of moles of CaCl2 = 0.0263 mol
mass of CaCl2 = nM
m = 0.0263 mol x 110,98 g/mol
m = 2.92 g
So the theoretical yield of CaCl2 = 2.92g
percentage yield = (2.50g/2.92)*100
percentage yield = 85.62%
