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.

In this exercise we have to use the knowledge of reaction to calculate the required percentage, in this way we find that:

The reaction equation is;

Since Al is in excess then NiSO4 is the limiting reactant. Now knowing that the data informed is:

From the reaction equation;

See more about reaction at link

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.

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%

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.