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22 January, 03:25

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

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Answers (1)
  1. 22 January, 03:39
    0
    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
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