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#?

1 Answer
Jan 28, 2017

Answer:

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.