Calculating Mass of Precipitate in a Chemical Reaction

[darkwatcher]

I have this lab in class, and i have not done chemsitry for a while..Can anyone help me with this...much thankful

purpose: To determine the mass of percipitate produced when 10ml of 0.20 mol/L carbonate reacts with 20 ml of 0.225 mol/L calcium chloride..

For the procedure i think i am ok

but the calculation is where i have no idea how to start..
any help would be helpful

thank you

 

[chemisttree]

Start with the precipitation reaction and then use the solubility product.

 

[darkwatcher]

Start with the precipitation reaction and then use the solubility product.

thank you for the reply..but I am still confused..i balanced out the equation which is
CaCl(2)(aq)+ Na(2)CO3(aq)-----> Ca(CO3)(l)+2NaCl(aq)

and then the moler mass are:
1) 10ml of 0.20 mol/l sodium carbonate is 0.01*0.20=0.002
2) 20ml of 0,225 mol/l calcium chloride is 0.020*0.225=0.0045


and the mass of the participate is 2.82g -2.60g(mass of the filter) = .72g

but now i have no idea where to go...i know i have to do stoichiometry but i am so confusedd

Plz anyone help

 

[chemisttree]

Now, what is the solubility product for CaCO3?

Correction: Your equation should be

CaCl(2)(aq)+ Na(2)CO3(aq)-----> Ca(CO3)(s)+2NaCl(aq)