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?

1 Answer
anor277
Jan 28, 2017

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.

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