Question

Given the following reaction:

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

How many moles of MgCl, will be produced from 32.0 g of Mg(OH)2, assuming HCl is available in excess?
moles (round to three significant figures)

Answer 1

Approximately
`\rm 0.549\; mol` (three significant figures.)

Step-by-step explanation:

Make sure that the reaction here is balanced. Look up a modern periodic table for the relative atomic mass data:

• Mg: 24.305.
• O: 15.999.
• H: 1.008.

Calculate the formula mass for
`\rm Mg(OH)_2`:

`\begin{aligned}& M\left(\mathrm{Mg(OH)_2}\right)\\ &= 24.305 + 2 * (15.999 + 1.008) \\ &= 58.319\; \rm g \cdot mol^(-1) \end{aligned}`.

Calculate the number of moles of formula units in that
`32.0\; \rm g` of
`\rm Mg(OH)_2`:

`\begin{aligned} & n \left(\mathrm{Mg(OH)_2}\right) \\ &= \frac{m \left(\mathrm{Mg(OH)_2}\right)}{M\left(\mathrm{Mg(OH)_2}\right)} \\ &= (32.0)/(58.319) \approx 0.548706\; \rm mol \end{aligned}`.

In the balanced equation for the reaction, the coefficient of
`\rm Mg(OH)_2` is the same as that of
`\rm MgCl_2`. In other words, for each mole of
`\rm Mg(OH)_2` formula units that this reaction consumes, one mole of
`\rm MgCl_2` will be produced. Therefore:

`\begin{aligned}& n\left(\mathrm{MgCl_2}\right)\\ &= n\left(\mathrm{Mg(OH)_2}\right) \\ &\approx 0.549\; \rm mol \end{aligned}`.

(Rounded to three significant figures.)