How many moles of iron can be recovered from 100.0 kg fe3o4?
2 answers:
Hello!
To know how many moles of iron can be recovered from 100 kg of Fe₃O₄ we'll need to use the molar mass of Fe₃O₄ and apply the conversion factor to go from kg of Fe₃O₄ to moles of Fe in the following way:
So, theoretically, one could recover 1192,68 moles of Fe from 100 kg of Fe₃O₄
Have a nice day!
To know how many moles of iron can be recovered from 100 kg of Fe₃O₄ we'll need to use the molar mass of Fe₃O₄ and apply the conversion factor to go from kg of Fe₃O₄ to moles of Fe in the following way:
So, theoretically, one could recover 1192,68 moles of Fe from 100 kg of Fe₃O₄
Have a nice day!
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Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:
Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:
Best regards.
Answer:
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
Explanation:
Let's consider the oxidation and reduction half-reactions and the global reaction.
Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻
Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)
Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red,cat - E°red,an
E° = 0.771 V - 0.154 V = 0.617 V
The Nernst equation allows us to calculate the cell potential (E) under the given conditions.
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.