chemistry Stoich

Question

85.00 g of calcium carbonate reacts with 45.00 g of iron(III) phosphate. What is mass, in grams, of iron(III) carbonate will form? 

please quick answer



Answer 1

Hi Sarah A. This is very much like the previous question that you posted, and that I answered. I will also answer this, but you really should try to follow the process and be able to perform these calculations on your own. Just a suggestion so you can pass the test without assistance.

Write a correctly balanced equation for the reaction taking place:

Calcium carbonate (MW = 100.0 g/mol) = CaCO₃

Iron(III) phosphate (MW = 150.8 g/mol) = FePO₄

Iron(III) carbonate (MW = 291.7 g/mol) = Fe₂(PO₄)₃

3CaCO₃ + 2FePO₄ ==> Fe₂(CO₃)₃ + Ca₃(PO4)₂ .. balanced equation

Next, since we are given amounts of BOTH reactants, we must find which is limiting:

CaCO₃: 85.00 g x 1 mol / 100.0 g) = 0.8500 moles (÷3->0.283)

FePO₄: 45.00 g x 1 mol / 150.8 g) = 0.2984 moles (÷2->0.149)

Since 0.149 is less than 0.283, FePO₄ is limiting

Finally, use the moles of limiting reactant to calculate mass of product, Fe₂(CO₃)₃ formed. Do this using stoichiometry of the balanced equation, molar mass of Fe₂(CO₃)₃ and dimensional analysis:

0.2984 mols FePO₄ x 1 mols Fe₂(CO₃)₃ / 2 mols FePO₄ x 291.7 g / mol = 43.52 g Fe₂(CO₃)₃

(NOTE: This is the theoretical yield, but Fe₂(CO₃)₃ is very unstable and probably would not exist to any significant extent)