What volume of 0.500m mgoh is required to neutralize 0.200l of 0.100m hcl?

Answer 1

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)

Answer 2

To neutralize HCl (acid) with Mg(OH)2 (base), we use the equation: HCl + Mg(OH)2 → MgCl2 + 2H2O. Since it is a 1:2 ratio, we need twice the volume of Mg(OH)2 compared to HCl. Therefore, you need 0.400 liters of 0.500 M Mg(OH)2 to neutralize 0.200 liters of 0.100 M HCl.

Answer 3

26mL

.01L*.47= .005

So... .18M*26= .005