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N = 5
V = 100 mL /1000 = 0.1 L
Peq = Mm / 2(OH)
Peq. = 74 g / mol = 37 g/ eq.
37 g ------ 1 eq-g
x ------ 5 eq-g
x = 185 g
185 g -------1 L
X -----0.1 L
X = 18.5 g de Ca(OH)2
la respuesta es la D