Help with Stoichiometry

I need help with this problem. I have chemistry finals tomorrow, and this is really confusing me.
Ca(NO3)2 + Na2CO3 -> CaCO3 +NaNO3
The question is "How much Calcium Nitrate and Sodium Carbonate are needed to produce .75g of the precipitate?"
I already balanced the equation (1-1-1-2) and I know that the precipitate is CaCO3. Besides that, I am completely lost. Any help would be greatly appreciated, thanks!

Given the information you already know, this is a really easy question!

First of all, you want 0.75g of CaCO3, so you first need to convert that to moles. The molar mass of CaCO3 is 100.089g/mol, so just divide 0.75g/100.08g/mol to get 0.00749mol CaCO3.

Since this is a one to one ratio equation, (We don't need to think about the Sodium Nitrate in this problem) we know that 0.00749mol of both Ca(NO3)2 and Na2CO3 were needed to produce 0.00749mol of CaCO3!

If you only need the answer in moles, then you've already solved the problem. However, most problems require the mass of the reactants.

To do this, just multiply the number of moles of each reagent by it's respective molar mass!

Na2CO3 = 0.794g needed
Ca(NO3)2 = 1.229g needed

Hope this answered your question!

hi,

first you have to understand which one is the precipitate. From your solubility rules you know that CaCO3 is the precipitate not NaNO3
then you can calculate the number of moles of CaCO3 there in .75 g

so number of moles of CaCO3 = .75 g * ( 1 mol / 100 g) ; 100 g is the MW of CaCO3
= .0075 moles of CaCO3

now you can use stoichiometry to calculate the amount of Calcium nitrate and sodium carbonate

so amount of sodium carbonate = .0075 moles of CaCO3 * ( 1 mole of Na2CO3 / 1 mole of CaCO3)
= .0075 moles of Na2CO3

so mass of Sodium Carbonate = .0075 moles of Na2CO3 * ( 105.99 g / 1 mole of Na2CO3)