Im assuming you mean Mg(OH)2 and not MgOH
The reaction between Mg(OH)2 and HCl is as follows: Mg(OH)2 (aq) + 2HCl (aq) -> MgCl2 (aq) + 2H2O (l)
First the number of mole is found: n(HCl) = c × v = 0.100M × 0.200L = 0.0200mol (to 3 significant figures)
Next we find the number of mole of Mg(OH)2:n(Mg(OH)2) ÷ n(HCl) = Coefficient of Mg(OH)2 ÷ Coefficient of HCl n(Mg(OH)2) ÷ n(HCl) = 1 ÷ 2 therefore:
n(Mg(OH)2) = (1 ÷ 2) × n(HCl) n(Mg(OH)2) = (1 ÷ 2) × 0.0200mol
n(Mg(OH)2) = 0.0100mol (to 3 significant figures)
Finally we calculate the volume of Mg(OH)2 reacted:
v(Mg(OH)2) = n ÷ c
v(Mg(OH)2) = 0.0100mol ÷ 0.500M
Therefore
v(Mg(OH)2) = 0.0200L (to 3 significant figures) = 20.0ml (to 3 significant figures)
26mL
.01L*.47= .005
So... .18M*26= .005