The number of electrons required to reduce 4. | vedclass

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Question

(B) The reduction reaction is: $Al^{3+} + 3e^- ightarrow Al$. Moles of $Al = \frac{	ext{Mass}}{	ext{Atomic mass}} = \frac{4.5 	imes 10^{-5} 	ext

Vedclass · Accepted Answer

$3.01 	imes 10^{18}$

Vedclass · Answer

(B) The reduction reaction is: $Al^{3+} + 3e^- ightarrow Al$. Moles of $Al = \frac{	ext{Mass}}{	ext{Atomic mass}} = \frac{4.5 	imes 10^{-5} 	ext{ g}}{27 	ext{ g/mol}} = 1.666 	imes 10^{-6} 	ext{ mol}$. Moles of electrons required $= 3 	imes 	ext{moles of } Al = 3 	imes 1.666 	imes 10^{-6} = 5 	imes 10^{-6} 	ext{ mol}$. Number of electrons $= 	ext{moles of electrons} 	imes N_A = 5 	imes 10^{-6} 	imes 6.022 	imes 10^{23} \approx 3.011 	imes 10^{18}$.

The number of electrons required to reduce $4.5 \times 10^{-5} \text{ g}$ of $Al^{3+}$ to $Al$ is:

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