What mass of gold is produced when 17.6^a of current are passed through a gold solution for 37.0 min ?

The mass of gold that is produced is 26.59 g

Using the formula

Where m is the mass

n is the number of equivalents

F is the Faraday constant ( F = 96485 C)

I is the current

and t is the time

From the given information

I = 17.6 A

t = 37.0 min = 37.0 × 60

t = 2220 secs

For gold

Atomic mass = 196.97 g/mol

and n = 3

Putting these parameters into the formula, we get

m = 26.59 g

Hence, the mass of gold that is produced is 26.59 g

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There is 26.58 grams of gold formed

Explanation:

Step 1: Data given

17.6 A of current are passed through a gold solution for 37.0 min

Molar mass of Au = 196.967 g/mol

Step 2: The equation

Au^3+ + 3e- → Au

Step 3: Calculate coulombs

17.6 Coulomb/s * 37.0 min * 60 sec/min = 39072 Coulombs

1 Faraday = 96500 Coulombs

Step 4: Calculate faraday

39072 Coulombs / 96500 Coulombs / Faraday = 0.40489 Faraday

Step 5: Calculate mass of gold formed

For every 3 Faraday of electricity used up , 1 mole Au is formed

0.40489 Faraday * 1 mole Au/ 3 Faraday = 0.13496 mole Au

196.967 g/mol * 0.13496 mol = 26.58 g Au

There is 26.58 grams of gold formed

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8