Question

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 10^4`g of `Mg(OH)_2`?

Answer 1

Approx. `23*kg` of `"magnesium metal.........."`

Explanation:

`Mg(s)+2H_2O(l) +Delta rarr Mg(OH)_2(s) + H_2(g)uarr`

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 `"assume"` that the reaction proceeds as written.

And thus moles of `"magnesium hydroxide"` `=` `(5.6xx10^4*g)/(58.32*g*mol^-1)` `=` `960.2*mol`

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 `"milk of magnesia"` (or so I just read). Given that the magenesium ion, `Mg^(2+)` is poorly absorbed by our intestinal tract, it tends to draw out water, and is thus used as a laxative.