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?

52.5 millimol LiAlH_(4) reacts with 15.6g (210 millimol) tert-butyl al

1.	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?
Answer» SOLUTION :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)`.