The question is incomplete, the complete question is;
In this stoichiometry problem, determine the percentage yield:
Excess aluminum metal reacts with aqueous nickel(II) sulfate to produce aqueous aluminum sulfate and nickel as a precipitate. In this reaction 108 g of aluminum were combined with 464 g of nickel(II) sulfate to produce 274 g of aluminum sulfate.
Answer:
80%
Explanation:
The reaction equation is;
2Al(s) + 3NiSO4(aq) --------> Al2(SO4)3 + 3Ni(s)
Since Al is in excess then NiSO4 is the limiting reactant.
Number of moles in 464 g of NiSO4 = mass/ molar mass
Molar mass of NiSO4 = 155 g/mol
Number of moles = 464g/155g/mol = 2.99 moles
Number of moles of Al2(SO4)3 = mass/molar mass
molar mass = 342 g/mol
Number of moles = 274g/342g/mol = 0.8 moles
From the reaction equation;
3 moles of NiSO4 yields 1 mole of Al2(SO4)3
2.99 moles of NiSO4 yields 2.99 * 1/3 = 1 mole of Al2(SO4)3
% yield = actual yield/ theoretical yield * 100/1
actual yield = 0.8 moles of Al2(SO4)3
Theoretical yield = 1 mole of Al2(SO4)3
% yield = 0.8/1 * 100 = 80%
In this exercise we have to use the knowledge of reaction to calculate the required percentage, in this way we find that:
[tex]80\%[/tex]
The reaction equation is;
[tex]2Al(s) + 3NiSO_4(aq) \rightarrow Al_2(SO_4)_3 + 3Ni(s)[/tex]
Since Al is in excess then NiSO4 is the limiting reactant. Now knowing that the data informed is:
From the reaction equation;
[tex]\% yield = actual\ yield/ theoretical\ yield * 100/1\\\% yield = 0.8/1 * 100 = 80%[/tex]
See more about reaction at brainly.com/question/3664113
