calculate moles of both reagents given and the moles of FeS that each of them would form if they were in excess
moles = mass / molar mass
moles Fe = 7.62 g / 55.85 g/mol
= 0.1364 moles
1 mole Fe produces 1 mole FeS
Therefore 7.62 g Fe can form 0.1364 moles FeS
moles S = 8.67 g / 32.07 g/mol
= 0.2703 moles S
1 mole S can from 1 moles FeS
So 8.67 g S can produce 0.2703 moles FeS
The limiting reagent is the one that produces the least product. So Fe is limiting.
The maximum amount of FeS possible is from complete reaction of all the limiting reagent.
We have already determined that the Fe can form up to 0.1364 moles of FeS, so this is max amount of FeS you can get.
Convert to mass
hope this helps :)
Answer: True
Explanation:
I just took the test and got it right
use the formula that was given to you
Answer : The balanced reduction half-reaction is:
![3Cu^{2+}+6e^-\rightarrow 3Cu](https://tex.z-dn.net/?f=3Cu%5E%7B2%2B%7D%2B6e%5E-%5Crightarrow%203Cu)
Explanation :
Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.
Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.
Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.
The given balanced redox reaction is :
![Al(s)+Cu^{2+}(aq)\rightarrow Al^{3+}(aq)+Cu(s)](https://tex.z-dn.net/?f=Al%28s%29%2BCu%5E%7B2%2B%7D%28aq%29%5Crightarrow%20Al%5E%7B3%2B%7D%28aq%29%2BCu%28s%29)
The half oxidation-reduction reactions are:
Oxidation reaction : ![Al\rightarrow Al^{3+}+3e^-](https://tex.z-dn.net/?f=Al%5Crightarrow%20Al%5E%7B3%2B%7D%2B3e%5E-)
Reduction reaction : ![Cu^{2+}+2e^-\rightarrow Cu](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%2B2e%5E-%5Crightarrow%20Cu)
In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 3 and then added both equation, we get the balanced redox reaction.
Oxidation reaction : ![2Al\rightarrow 2Al^{3+}+6e^-](https://tex.z-dn.net/?f=2Al%5Crightarrow%202Al%5E%7B3%2B%7D%2B6e%5E-)
Reduction reaction : ![3Cu^{2+}+6e^-\rightarrow 3Cu](https://tex.z-dn.net/?f=3Cu%5E%7B2%2B%7D%2B6e%5E-%5Crightarrow%203Cu)
The balanced redox reaction will be:
![2Al(s)+3Cu^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Cu(s)](https://tex.z-dn.net/?f=2Al%28s%29%2B3Cu%5E%7B2%2B%7D%28aq%29%5Crightarrow%202Al%5E%7B3%2B%7D%28aq%29%2B3Cu%28s%29)
Thus, the balanced reduction half-reaction is:
![3Cu^{2+}+6e^-\rightarrow 3Cu](https://tex.z-dn.net/?f=3Cu%5E%7B2%2B%7D%2B6e%5E-%5Crightarrow%203Cu)