## Chemistry: The Central Science (13th Edition)

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: $C = \frac{n(moles)}{V(L)}$ $0.0128 = \frac{0.0984}{V(L)}$ $V(L) = \frac{0.0984}{0.128}$ $V(L) = 0.76875 L$