Answer:
Approximately [tex]\rm 0.549\; mol[/tex] (three significant figures.)
Explanation:
Make sure that the reaction here is balanced. Look up a modern periodic table for the relative atomic mass data:
Calculate the formula mass for [tex]\rm Mg(OH)_2[/tex]:
[tex]\begin{aligned}& M\left(\mathrm{Mg(OH)_2}\right)\\ &= 24.305 + 2 \times (15.999 + 1.008) \\ &= 58.319\; \rm g \cdot mol^{-1} \end{aligned}[/tex].
Calculate the number of moles of formula units in that [tex]32.0\; \rm g[/tex] of [tex]\rm Mg(OH)_2[/tex]:
[tex]\begin{aligned} & n \left(\mathrm{Mg(OH)_2}\right) \\ &= \frac{m \left(\mathrm{Mg(OH)_2}\right)}{M\left(\mathrm{Mg(OH)_2}\right)} \\ &= \frac{32.0}{58.319} \approx 0.548706\; \rm mol \end{aligned}[/tex].
In the balanced equation for the reaction, the coefficient of [tex]\rm Mg(OH)_2[/tex] is the same as that of [tex]\rm MgCl_2[/tex]. In other words, for each mole of [tex]\rm Mg(OH)_2[/tex] formula units that this reaction consumes, one mole of [tex]\rm MgCl_2[/tex] will be produced. Therefore:
[tex]\begin{aligned}& n\left(\mathrm{MgCl_2}\right)\\ &= n\left(\mathrm{Mg(OH)_2}\right) \\ &\approx 0.549\; \rm mol \end{aligned}[/tex].
(Rounded to three significant figures.)
