The decomposition of Al2O3 to form pure aluminum (Al) along with the release of oxygen (O2) gas can be represented by the following chemical reaction:
2Al2O3 → 4Al + 3O2
Based on the reaction stoichiometry:
2 moles of aluminum oxide decomposes to form 4 moles of pure aluminum
Therefore, if 29.0 moles of aluminum is recovered, then the moles of aluminum oxide required will be:
= 29.0 moles Al * 2 moles Al2O3/4 moles Al
= 14.5 moles of Al2O3
Answer:
132.17 g
Explanation:
The reaction given , in the question is -
CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S ( g )
From the reaction , 4 mole of S is required for the production of 1 mole of CS₂ .
since ,
Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂
Since ,
the Molecular weight of CS₂ = 76
Given , mass of CS₂ = 72.57 g
Moles of CS₂ = 72.57 / 76 = 0.95 mol
Since ,
The yield is 92.0 % .
Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles
Mass of S required = 4.13 * 32 = 132.17 g .
<u>Answer:</u> Electrons are taken up by
and they are lost by 
<u>Explanation:</u>
Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously. It is also called the reaction where the exchange of electrons takes place.
An oxidation reaction is defined as the reaction in which a chemical species loses electrons takes place. In this reaction, the oxidation state of a substance gets increased.
A reduction reaction is defined as the reaction in which a chemical species gains electrons takes place. In this reaction, the oxidation state of a substance gets reduced.
For the given chemical reaction:

The half-reactions for this redox rection follows:
<u>Oxidation half-reaction:</u> 
<u>Reduction half-reaction:</u> 
Hence, electrons are taken up by
and they are lost by 