Respuesta :
Answer:
The limiting reactant is CuSO₄.
Explanation:
The reaction is:
Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s) (1)
To find the limiting reactant we need to find the number of moles of the reactants.
[tex] \eta_{Fe} = \frac{m}{A_{r}} [/tex]
Where:
m: is the mass of iron = 3.26 g
[tex]A_{r}[/tex]: is the standard atomic weight of iron = 55.845 g/mol
[tex]\eta_{Fe} = \frac{3.26 g}{55.845 g/mol} = 0.058 moles [/tex]
[tex] \eta_{CuSO_{4}} = M*V [/tex]
Where:
M: is the concentration of the CuSO₄ = 0.200 mol/dm⁻³
V: is the volume of the solution = 80.0 cm³
First, we need to convert the units of the volume to dm³ knowing that 1 dm = 10 cm and 1 L= 1 dm³.
[tex] V = 80.0 cm^{3}*\frac{1 dm^{3}}{(10 cm)^{3}} = 0.080 dm^{3} [/tex]
Now, the number of CuSO₄ moles is:
[tex] \eta_{CuSO_{4}} = M*V = 0.200 mol/dm^{3}*0.080 dm^{3} = 0.016 moles [/tex]
So, to determine the limiting reactant we need to use the molar ratio from equation (1), Fe:CuSO₄ = 1:1
[tex]\eta_{Fe} = \frac{1 mol Fe}{1 mol CuSO_{4}}*0.016 moles \: CuSO_{4} = 0.016 moles[/tex]
Since we need 0.016 moles of Fe to react with 0.016 moles of CuSO₄ and initially we have 0.058 moles of Fe, then the limiting reactant is CuSO₄.
Therefore, the limiting reactant is CuSO₄.
I hope it helps you!
The limiting reactant would be CuSO4
Looking at the equation of the reaction:
Fe (s) + CuSO4 (aq) → FeSO4 (aq) + Cu (s)
The mole ratio of Fe to CuSO4 is 1:1. Thus, every 1 mole of Fe will require 1 mole of CuSO4 for reaction.
Recall that: mole = mass/molar mass
Also: mole = molarity x volume
mole of Fe = 3.26/55.8
= 0.058 moles
mole of CuSO4 = 0.200 x 80/1000
= 0.016
The mole of Fe in the reaction is more than the mole of CuSO4. Thus, CuSO4 is the limiting reactant of the reaction. Fe is in excess.
More on limiting reactants can be found here: https://brainly.com/question/14225536?referrer=searchResults