What mass of silica containing 10% impurities is required to interact with 150 g of limestone containing 20% impurities?

Answer: m (silica) = 80 g.
Explanation of the solution to the problem: to solve this problem, you first need to write the equation of the chemical reaction: SiO2 + CaCO3 = CaSiO3 + CO2.
We find the mass of CaCO3, for this we multiply the mass of limestone by the mass fraction of pure calcium carbonate in it:
m (CaCO3) = 150 * 80/100 = 120 g.
The amount of CaCO3 substance is equal to:
n (CaCO3) = 120/100 = 1.2 mol.
n (SiO2) = n (CaCO3) = 1.2 mol, since they react in a 1: 1 ratio.
Find the mass of pure SiO2:
m (SiO2) = n (SiO2) * M = 1.2 * 60 = 72 g.
The mass of silica (SiO2 with impurities) is:
m (silica) = m (SiO2) * 100% / 90% = 72 * 100% / 90% = 80 g.