In this stoichiometry problem, determine the limiting reactant.

a. Aluminum metal reacts with aqueous nickel(II)

b. sulfate to produce aqueous aluminum sulfate and nickel as a precipitate.

c. In this reaction 108 g of aluminum were combined with 464 g of nickel(II)

d. sulfate to produce 274 g of aluminum sulfate.

Limiting reactant is NiSO₄

Explanation:

The reaction of aluminum metal with aqueous nickel(II)  sulfate to produce aqueous aluminum sulfate and nickel is:

2 Al(s) + 3 NiSO₄ → Al₂(SO₄)₃ + 3 Ni  

That means 2 moles of Al react with 3 moles of nickel sulfate.

Moles of Al and NiSO₄ are:

Al: 108g × (1mol / 26.98g) = 4.00 moles of Al

NiSO₄: 464g × (1mol / 154.75g) = 3.00 moles of NiSO₄

For a complete reaction of aluminium there are necessary:

4.00mol Al ₓ ( 3 moles NiSO₄ / 2 moles Al) = 6 moles of NiSO₄

As you have just 3.00 moles of NiSO₄, the limiting reactant is NiSO₄

the answer for sodium is 0.186

the answer for rubidium is 0.251