70.0 g. The decomposition of 125 g CaCO3 produces 700 g CaO.
MM = 100.09 56.08
CaCO3 → CaO + CO2
Mass 125 g
a) Moles of CaCO3 = 125 g CaCO3 x (1 mol CaCO3/100.09 g CaCO3)
= 1.249 mol CaCO3
b) Moles of CaO = 1.249 mol CaCO3 x (1 mol CaO/1 mol CaCO3)
= 1.249 mol CaO
c) Mass of CaO = 1.249 mol CaO x (56.08 g CaO/1 mol CaO) = 70.0 g
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HClO <-----> H+ + ClO- at equilibrium [HClO]= 0.015-x [H+] = [ClO-]= x 3.0 x 10^-8 = x^2/ 0.015-x x = 2.1 x 10^-5 % ionization = 2.1 x 10^-5 x 100/ 0.015=0.14
