Question

What volume of 0.140 HCl is needed to neutralize 2.58 of Mg(OH)2
?

Answer 1

The units of measurement are missing. Is that 0.140g of HCl and 2.58g of Mg(OH)2?

Answer 2

The following calculations are based on the assumption that 0.140 is Molarity.

Answer 3

\[2.58g Mg(OH)_{2} * (1 mol Mg(OH)_{2})/(58.3196g Mg(OH)_{2}) = 0.042 mol Mg(OH)_{2}\]
The balanced formula for neutralization should be:
\[Mg(OH)_{2}(aq) + 2HCl(aq) \rightarrow 2H _{2}O + Mg(aq) + Cl(aq)\]
based on this we can calculate the following:
\[0.0442 mol Mg(OH)_{2}*(2 mol HCl)/(1 mol Mg(OH)_{2})=0.0885 mol HCl\]
\[0.140M HCl = 0.140 (mol/L) HCl\]
\[V=(0.0885 mol HCl)/(0.140 (mol/L) HCl) = 0.632 L HCl\]