A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al (s) + 3CuSO4 (aq) → Al2 (SO4) 3 (aq) + 3Cu (s)

Aluminum

Copper

Copper (II) sulfate

Aluminum sulfate

Copper (II) sulfate

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