Answer:
Copper (II) sulfate
Step-by-step explanation:
Given reaction is
2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)
Amount of aluminum = 1·25 g
Amount of copper (II) sulfate = 3·28 g
Atomic weight of Al = 26 g
Molecular weight of CuSO4 ≈ 159·5
Number of moles of Al = 1·25 ÷ 26 = 0·048
Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021
From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required
So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required
For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required
∴ Number of moles of copper (ll) sulfate required = 0·072
But we have only 0·021 moles of copper (ll) sulfate
As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate
∴ The limiting reactant is copper (ll) sulfate