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;

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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%

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"