Can someone help me with this question about a chemical reaction?

Here's the chemical reaction: 2Al + 3 CuSo4 = 1 Al2(SO4)3 + 3Cu

Here's the question.

If you have 10.98 grams of copper (II) sulfate and want to react it ALL, how many grams of Al would you need? How many grams of each product (Al2(SO4)3 and Cu) would you produce?

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  • 7 years ago
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    2 Al + 3 CuSO4 → Al2(SO4)3 + 3 Cu

    (10.98 g CuSO4) / (159.6095 g CuSO4/mol) x (2 mol Al / 3 mol CuSO4) x (26.98154 g Al/mol) = 1.237 g Al

    (10.98 g CuSO4) / (159.6095 g CuSO4/mol) x (1 mol Al2(SO4)3 / 3 mol CuSO4) x (342.1527 g Al2(SO4)3/mol) =

    7.846 g Al2(SO4)3

    (10.98 g CuSO4) / (159.6095 g CuSO4/mol) x (3 mol Cu / 3 mol CuSO4) x (63.5463 g Cu/mol) = 4.372 g Cu

  • 7 years ago

    First you have correctly written the balanced reaction eq'n.

    Next we need to calculate the moles of copper sulphate that reacted.

    To do this use the eq'n

    moles = mas(g) / Mr

    The Mr ( Relative Molecular mass) is calculated by : -

    Cu x 1 = 63.5 x 1 = 63.5

    S x 1 = 32 x 1 = 32

    O x 4 = 16 x 4 = 64

    63.5 + 32 + 64 = 159.5

    Hence mol(CuSO4) = 10.98 / 159.5 = 0.06884 ( This is equivalent to '3' in the reaction eq'n above).

    mol(Al) = 0.06884 x 2 /3 = 0.04589this equivalent to '2' in the reaction eq'n above).

    mass(Al(g)) = moles x Mr(Al)

    mass(Al) = 0.04589 x 27 = 1.239 g is the required mass.

    The total reactant masses is 10.98 + 1.239 = 12.219 g

    So by the Law of Conservation of Mass, the product masses MUST be the same total.

    We have 0.06884 ( This is equivalent to '3' in the reaction eq'n above). of product Cu.

    Hence mass (Cu) = 0.06884 x 63.5 = 4.371 g

    Hence mass of (Al2(SO4)3 = 12.219 - 4.371 = 7.848 g

    Hope that helps!!!!

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