When 412.5g of calcium carbonate react with 521.9g of aluminum fluoride, how many grams of each product can be produced

Answer;

321.8 g CaF2

321.5 g Al2 (CO3) 3

Explanation;

The equation for the reaction is;

3 CaCO3 + 2 AlF3 → 3 CaF2 + Al2 (CO3) 3

Number of moles of CaCO3 will be;

= (412.5 g CaCO3) / (100.0875 g CaCO3/mol)

= 4.12139 mol CaCO3

Number of moles of AlF3 will be;

= (521.9 g AlF3) / (83.9767 g AlF3/mol)

= 6.21482 mol AlF3

But;

4.12139 moles of CaCO3 would react completely with 4.12139 x (2/3) = 2.74759 moles of AlF3.

Thus; there is more AlF3 present than that, so AlF3 is in excess, and CaCO3 is the limiting reactant.

Therefore;

Mass of CaF2 will be;

(4.12139 mol CaCO3) x (3/3) x (78.0752 g CaF2/mol) = 321.8 g CaF2

Mass of Al2 (CO3) 3 on the other hand will be;

(4.12139 mol CaCO3) x (1/3) x (233.9903 g Al2 (CO3) 3/mol) = 321.5 g Al2 (CO3) 3