a)MCaO=40+16=56 (g)
%Ca=\(\frac{40}{56}\).100\(\approx\)71,429%
%O=\(\frac{16}{56}\).100\(\approx\)28,571%
b)MCaCO3=40+12+3.16=100 (g)
%Ca=\(\frac{40}{100}\).100=40%
%C=\(\frac{12}{100}\).100=12%
%O=\(\frac{3.16}{100}\).100=48%
c)MCa(OH)2=40+2(16+1)=74 (g)
%Ca=\(\frac{40}{74}\).100\(\approx\)54,054%
%O=\(\frac{2.16}{74}\).100\(\approx\)43,243%
%Ca=\(\frac{2.1}{74}\).100\(\approx\)2,703%