In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten c
1 answer:
Answer:
0.382g
Explanation:
Step 1: Write the reduction half-reaction
Al³⁺(aq) + 3 e⁻ ⇒ Al(s)
Step 2: Calculate the mass of Al produced when a current of 100. A passes through the cell for 41.0 s
We will use the following relationships.
- 1 A = 1 C/s
- 1 mole of electrons has a charge of 96486 C (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons pass through the cell.
- The molar mass of Al is 26.98 g/mol.
The mass of Al produced is:
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Answer:
Two moles.
Explanation:
Sulphuric (sulfuric) acid is a diprotic acid. When one mole of
molecules dissolve in water, two moles of
ions would be produced.
.
On the other hand, sodium hydroxide is a monoprotic base. When one mole of
formula units dissolve in water, only one mole of hydroxide ions
would be produced.
.
Note that and
react at a one-to-one ratio:
.
As a result, it would take of
to react with the
of
that was released when
of
is dissolved in water. Since one mole of
formula units could produce only one mole of
, it would take
of
formula units to produce that
of
for reacting with
of
.