During the combustion of 40 g of technical magnesium containing 4% impurities, 36 g of magnesium

During the combustion of 40 g of technical magnesium containing 4% impurities, 36 g of magnesium oxide were formed. Calculate the yield fraction of what is theoretically possible.

MgCO3-> MgO + CO2
Find the mass of pure MgCO3 m (MgCO3) pure = m (MgCO3) * w impurities = 50g * 0.16 = 8g
Now let’s find the amount of substance in MgCO3 n (MgCO3) = m (MgCO3) / M (MgCO3) = 8 g / 84 g / mol = 0.095 mol
According to the reaction equation n (MgCO3) = n (MgO) = 0.095 mol
Well, let’s find the mass of the formed magnesium oxide m (MgO) = 0.095 mol * 40 g / mol = 3.8 g