How do you balance #LiAl(OH)_4 + H_2O -> LiOH + Al(OH)_3 + H_2O#?

2 Answers
Jan 27, 2016

It's already balanced. Here, let's see what happens when we check what's on each side.

#"LiAl"("OH")_4 + cancel("H"_2"O") -> "LiOH" + "Al"("OH")_3 + cancel("H"_2"O")#

#color(green)("Li")color(highlight)("Al")color(blue)(("OH")_4) -> color(green)("Li")color(blue)("OH") + color(highlight)("Al")color(blue)(("OH")_3)#

Yep, it's fine the way it is. One lithium on each side, one aluminum on each side, and four #"OH"# groups on each side.

As for the charges, #"Al"("OH")_4^(-)# is balanced by #"Li"^(+)#, #"OH"^(-)# is balanced by #"Li"^(+)#, and #3xx"OH"^(-)# is balanced by #"Al"^(3+)#.

Jan 27, 2016

Answer:

Do you mean the reaction of lithium tetrahydroaluminate (#LiAlH_4#) with water?

Explanation:

#LiAlH_4(s) + 4H_2O(l) rarr LiAl(OH)_4(aq) + 4H_2(g)uarr#

Lithium tetrahydroaluminate is an important hydride transfer reagent, and a very common reductant in organic chemistry. The reduction reaction is performed in THF (#C_4H_8O#) or ether (both solvent must be dry).

#LiAlH_4# can transfer up to 4 hydrides (with commercial grades usually it transfers about 3). When your organic reduction is finished, the mixture is worked up with water. This can get pretty violent (especially if the lithal has been added 1:1, as is quite commonly done).