For the complete reduction of 5.8 g of acetone to isopropyl alcohol, the quantity of LiAlH₄ required (assuming chemical yield to be 100%) is approximately [mass: Li = 6.9, Al = 27]

The reaction will be:

Acetone + LiAlH₄ = isopropyl alcohol

From the reaction, we can conclude that 1 mole of acetone with 1 mole of LiAlH₄ gives 1 mole of isopropyl alcohol.

Total moles of acetone = $\frac{\mathrm{given} \mathrm{weight} \mathrm{of} \mathrm{acetone}}{\mathrm{molecular} \mathrm{weight}}$

= $\frac{5.8}{58}$ = $\frac{1}{10}$

Since the number of moles of LiAlH₄ is equal to the moles of acetone, therefore, number of moles of LiAlH₄ will be $\frac{1}{10}$.

Since, moles = $\frac{\mathrm{weight}}{\mathrm{molecular} \mathrm{weight}}$

Molecular weight of LiAlH₄ = 6.9 + 27+ (4 x 1)

$\frac{1}{10}$ = $\frac{\mathrm{weight}}{\left(6.9 + 27 + 4\right)}$

weight = $\frac{37.9}{10}$

weight = 3.79 g

Therefore, the quantity of LiAlH₄ required is 3.79 g.