Answer
0.76875L
Work Step by Step
1. Convert 2.87g of $Mg(OH)_2$ to nº of moles:
Molar Mass: $Mg(OH)_2$:
Mg: 24.3 * 1
O: 16 * 2
H: 1 * 2
24.3 + 32 + 2 = 58.3 g/mol
$mm = \frac{mass(g)}{n(moles)}$
$58.3 = \frac{2.87}{n(moles)}$
$n(moles) = \frac{2.87}{58.3}$
$n(moles) = 0.0492$
2. Since the proportion is 2 $HCl$ to 1 $Mg(OH)_2$:
$n(moles)(HCl) = 2 \times 0.0492$
$n(moles)(HCl) = 0.0984$
3. Find the volume, using the nº of moles and the concentration:
$C = \frac{n(moles)}{V(L)}$
$0.0128 = \frac{0.0984}{V(L)}$
$V(L) = \frac{0.0984}{0.128}$
$V(L) = 0.76875 L$
