How many kilograms of CaCO3 should be added for every kilogram of iron ore processed in a blast furncace...?

Question

if the limestone is 98.20% pure and the iron ore contains 12.80% SiO2 by mass? (Note: SiO2 (s) + CaCO3 (s) ---> CaSiO3 (s) + CO2 (g) )

Answer 1

We consider 1000 g of this compound.
128.0 g of SiO2 ==> 128.0 g / 60.086 = 2.13 mol = mol CaCO3
Molecular weight CaCO3 = 100 g /mol
2.13 mol x 100 = 213 g CaCO3 react with 128 g of SiO2
1000 - 12.80 = 987.2 g of Iron ore
1000 g ( iron ore ) : x = 987.2 : 213
x = 215.7 g of CaCO3 needed for every Kg of iron ore

Answer 2

3.5