Determine the theoretical yield. aluminum metal reacts with aqueous nickel II sulfate to produce aqueous aluminum sulfate and nickel as a precipitate. in this reaction 108 grams of aluminum were combined with 464 grams of nickel II sulfate to produce 274 grams of aluminum sulfate.

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 correct equation for this reaction is;

2Al(s) + 3NiSO4(aq) > Al2(SO4)3 + 3Ni(s)

We already know that Al is in excess then NiSO4 is the limiting reactant.

Hence, 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

So, Number of moles = 274g/342g/mol = 0.8 moles

Given the stoichiometric 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

percentage yield is given by; actual yield/ theoretical yield * 100/1

actual yield =  0.8 moles of Al2(SO4)3

Theoretical yield = 1 mole of Al2(SO4)3

Hence;

% yield = 0.8/1 * 100 = 80%

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation:

lol idk man i guess u gotta ask bill nye the science guy