52.5 millimol `LiAlH_(4)` reacts with 15.6g (210 millimol) tert-butyl alcohol. In the following reaction 157.5 millimol `H_(2)` is produced.
`LiAlH_(4)+3(CH_(3))_(3)COH to 3H_(2)+LiC(CH_(3))_(3) O]_(3)AlH`
On adding extra methanol or alchol in the above reaction, displacement of the 4th hydrogen atom `LiAlH_(4) ` will be observed by the following reaction.
`Li[(CH_(3))_(3)O]_(3)AlH+CH_(3)OH to H_(2)+Li(CH_(3))_(3)O]_(3)[CH_(3)]Al`
How much `H_(2)` will evolve on adding methanol?

Explanation

1 mol of `H_(2)` will be produced from 1 mol of `Li[(CH_(3))_(3)CO]_(3)AlH`.
1 mol of `Li[(CH_(3))_(3)O]_(3)` AlH will be produced from 1 mol of `LiAlH_(4)`. From second equation, 1 mol of `Li[(CH_(3))_(3)CO]_(3)AlH` will produce 1 mol of `H_(2)`. 52.5 millimol of `Li[(CH_(3))_(3)CO]_(3)AlH` will react with excess of `CH_(4)` to produce 52.5 millimon of `H_(2)`.