Answer:
0.453 moles
Explanation:
The balanced equation for the reaction is:
2Fe(s) + 3O2(g) ==> 2Fe2O3
From the equation, mass of O2 involved = 16 x 2 x 3 = 96g
mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2
= 100g
Therefore 96g of O2 produced 100g of Fe2O3
32.2g of O2 Will produce 100x32.2/96
= 33.54g of Fe2O3
Converting it to mole using number of mole = mass/molar mass
but molar mass of Fe2O3 = 26 + (16 X 3)
= 74g/mole
Therefore number of mole of 33.54g of Fe2O3 = 33.54/74
= 0.453 moles
3 cubic centimeters, Volume= density/ mass
Answer:
Cathode: Mn → Mn²⁺ + 2e⁻ (Oxidation)
Anode: Zn²⁺ + 2e⁻ → Zn (Reduction)
Mn | Mn²⁺ || Zn²⁺ | Zn
Explanation:
To identify the half reaction we need to see the oxidation states.
Mn(s) → Ground state → Oxidation state = 0
Mn(NO₃)₂ → Mn²⁺ → The oxidation state has increased.
This is the oxidation reaction. It has released two electrons:
Mn → Mn²⁺ + 2e⁻
Zn(NO₃)₂ → Zn²⁺
Zn → Ground state → The oxidation state was decreased.
This is the reduction reaction. It has gained two electrons:
Zn²⁺ + 2e⁻ → Zn
Cathode: Mn → Mn²⁺ + 2e⁻
Anode: Zn²⁺ + 2e⁻ → Zn
Mn | Mn²⁺ || Zn²⁺ | Zn
Answer:
The amswer to the question is A
