Calculating the number of Moles formed of a Precipitate.

When aqeuous solutions of Potassium Carbonate K2CO3 and Iron(III) Nitrate Fe(NO3)3 are mixed, a precipitate of Iron (III) Carbonate Fe2(CO3)3 is formed. In a beaker, 556.8mL (0.5568L) of 0.2540M Potassium Carbonate is added to an excess amount of Iron (III) Nitrate. Assuming the precipitation goes to completion, calculate the chemical amount (moles) of Fe2(CO3)3 formed in the beaker.


The answer given is 0.04714mol of Fe2(CO3)3 is formed in the beaker. I cannot understand how they got to this answer and this question is in my mid-semester test so I need to know how to do it correctly, please explain and show your working in each step so I can learn HOW to do it, thanks!

I believe you need to understand the Stoichiometry ratio in order to solve the question but I'm getting confused, I have an equation of: 2Fe^3+ + 3CO3^2- = 1 Fe2(CO3)3. Hence this ratio I believe is 2 moles of Iron react with 3 moles of Carbonate to form 1 mole of Iron (III) Carbonate.

If somebody can show me how this question is answered to get the final answer of 0.04714mol I would be really appreciative!!!

Thanks