Respuesta :
Explanation:
The given chemical reaction will be as follows.
[tex]Al(NO_{3})_{3} \rightarrow Al(s) + 3e^{-}[/tex]
Also, mass deposited can be calculated using the formula as follows.
W = Zit
where, W = weight or mass of the substance
Z = electrochemical equivalent
i = current
t = time in seconds
Calculate value of Z for the given reaction as follows.
Z = [tex]\frac{\text{molar mass of Al}}{\text{no. of electrons} \times 96500}[/tex]
molar mass of Al = 26.98 g/mol
Z = [tex]\frac{26.98 g/mol}{3 \times 96500}[/tex]
= [tex]9.32 \times 10^{-5}[/tex] g/mol
Therefore, putting the given values into the above formula as follows.
W = Zit
= [tex]9.32 g/mol \times 10^{-5} \times 0.834 amp \times 1.175 \times 3600 sec[/tex]
= 0.328 g
Thus, we ca conclude that 0.328 g of aluminium can be electroplated in the given situation.
