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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» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :1 mol of `H_(2)` will be produced from 1 mol of `Li[(CH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))_(3)CO]_(3)AlH`. <br/> 1 mol of `Li[(CH_(3))_(3)O]_(3)` AlH will be produced from 1 mol of `LiAlH_(4)`. From second <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a>, 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 <a href="https://interviewquestions.tuteehub.com/tag/react-613674" style="font-weight:bold;" target="_blank" title="Click to know more about REACT">REACT</a> with excess of `CH_(4)` to produce 52.5 millimon of `H_(2)`.</body></html>


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