Question

In: Chemistry

How many moles of zinc are in 4.80 kg of zinc?

± Introduction to Electroplating

Learning Goal:

To relate current, time, charge, and mass for electroplating calculations.

Electroplating is a form of electrolysis in which a metal is deposited on the surface of another metal. To quantify electrolysis, use the following relationships.

Electric current is measured in amperes (A), which expresses the amount of charge, in coulombs (C), that flows per second (s):

1 A=1 C/s

Another unit of charge is the faraday (F), which is equal to a mole of electrons and is related to charge in coulombs as follows:

1 F=1 mol e−=96,500 C.

Galvanized nails are iron nails that have been plated with zinc to prevent rusting. The relevant reaction is

Zn2+(aq)+2e−→Zn(s)

For a large batch of nails, a manufacturer needs to plate a total zinc mass of 4.80 kgon the surface to get adequate coverage.

Part A

How many moles of zinc are in 4.80 kg of zinc?

Express your answer to three significant figures and include the appropriate units.

73.4 mol (correct answer)

Part B How many coulombs of charge are needed to produce 73.4 mol of solid zinc?

Express your answer to three significant figures and include the appropriate units.

Solutions

Expert Solution

 

Number of moles (n) is given by, \(\mathrm{n}=\frac{\mathrm{w}}{\mathrm{M}}\) ...(i)

Where, \(\mathrm{w}=\) weight of substance \(\mathrm{M}=\mathrm{M}\) olar \(\mathrm{m}\) as \(\mathrm{s}\)

We know that, \(\mathrm{w}=4.80 \mathrm{Kg}=4.80 \times 1000 \mathrm{~g}=4800 \mathrm{~g}[\mathrm{Given}]\)

\(\mathrm{M}=65.38 \mathrm{~g}\)

Substituting the se values in equation (i), we have \(\mathrm{n}=\frac{4800 \mathrm{~g}}{65.38 \mathrm{~g} \mathrm{~mol}^{-1}}\)

\(=73.4 \mathrm{~m} \circ 1\)

Hence, number of moles is \(73.4 \mathrm{~mol}\).

 

We know that, \(\mathrm{Zn}^{2+}+2 e^{-} \longrightarrow \longrightarrow \mathrm{Zn}\)

Since, \(1 \mathrm{~F}=1 \mathrm{~mol} \mathrm{e}^{-} \mathrm{s} 0,\) for \(2 \mathrm{e}^{-}, 2 \mathrm{~F}\) of electricity is required to produce \(1 \mathrm{~m} \circ 1\) of \(\mathrm{Zn}\) \(1 \mathrm{~F}=96,500 \mathrm{C}\)

or, \(2 \mathrm{~F}=2 \times 96,500 \mathrm{C}=1,93,000 \mathrm{C}\)

Since, 1 mol of \(Z\) n is produced by 1,93,000 C of charge

=1,42,00,000 C

Hence, the coulom bs of charge required to produce \(73.4 \mathrm{~m}\) ol of solid zinc is \(1,42,00,000 \mathrm{C}\)

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