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Three moles of `B_(2)H_(6)` are completely reacted with methanol. The number of moles of boron containing product formed is:

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Three moles of `B_(2)H_(6)` are completely reacted with methanol. The number of moles of boron containing product formed is:
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`B_(2)H_(6)+6CH_(3)OH to 2B(OCH_(3))_(3)+6H_(2)O`
`"3 mol "B_(2)H_(6)"will give 6 moles" B(OCH_(3))_(3)`
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