Perform a solution:
Under the condition of the problem, we will make the equation of the calcination process:
Caco3 = Cao + H2O – decomposition, obtained calcium oxide (forgery);
M (Caco3) = 100 g / mol;
M (CaO) = 56 g / mol;
M (Caco3) = 10,000 * (1 – 0.10) = 9000 g = 9 kg (weight without impurities).
We define the amount of starting material:
Y (Caco3) = m / m = 9000/100 = 90 mole;
Y (Cao) = 90 mole Since the amount of substances is 1 mol.
We find a lot of product:
M (CaO) = y * m = 90 * 56 = 2040 g
Answer: Mass of the oversized lime is 2040 g
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