# In the following reaction: Mg(s) + 2H_2O(l) -> Mg(OH)_2(s) + H_2(g). What mass of Mg must you use to make 5.6 x 10^4g of Mg(OH)_2?

Jan 28, 2017

Approx. $23 \cdot k g$ of $\text{magnesium metal..........}$

#### Explanation:

$M g \left(s\right) + 2 {H}_{2} O \left(l\right) + \Delta \rightarrow M g {\left(O H\right)}_{2} \left(s\right) + {H}_{2} \left(g\right) \uparrow$

The given equation correctly identifies the molar equivalence of reactants and products. One mole of metal reduces two moles of water, and yields PRECISELY 1 mole of metal hydroxide. In practice, you would need a fair bit of heat (and activated metal) to make this go. We $\text{assume}$ that the reaction proceeds as written.

And thus moles of $\text{magnesium hydroxide}$ $=$ $\frac{5.6 \times {10}^{4} \cdot g}{58.32 \cdot g \cdot m o {l}^{-} 1}$ $=$ $960.2 \cdot m o l$

And so we need the following mass of metal:

960.2*cancel(mol)xx24.3*g*cancel(mol^-1)=??*g

Magnesium hydroxide is part of our pharmacopeia, and is administered as $\text{milk of magnesia}$ (or so I just read). Given that the magenesium ion, $M {g}^{2 +}$ is poorly absorbed by our intestinal tract, it tends to draw out water, and is thus used as a laxative.