What volume of .500 M HCl is needed to neutralize .287g Mg(OH)2?

Answer 1

To find the volume of HCl needed to neutralize Mg(OH)2, we first need to balance the chemical equation of the reaction between HCl and Mg(OH)2:

2HCl + Mg(OH)2 -> MgCl2 + 2H2O

From the balanced equation, we can see that it takes 2 moles of HCl to react with 1 mole of Mg(OH)2.

Now let's calculate the number of moles of Mg(OH)2:

molar mass of Mg(OH)2 = 24.31 g/mol + 16.00 g/mol + 2*(1.01 g/mol) = 58.33 g/mol

moles of Mg(OH)2 = mass / molar mass = 0.287 g / 58.33 g/mol = 0.00492 mol

Since it takes 2 moles of HCl to react with 1 mole of Mg(OH)2, we need 2 * 0.00492 mol = 0.00984 mol of HCl.

Now, let's calculate the volume of 0.500 M HCl needed:

molarity (M) = moles / volume (L)

0.500 M = 0.00984 mol / volume

volume = 0.00984 mol / 0.500 mol/L = 0.0197 L = 19.7 mL

Therefore, 19.7 mL of 0.500 M HCl is needed to neutralize 0.287 g of Mg(OH)2.