Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

Theoretical Yield?

Determine the theoretical yield of alum (MM=474.38), to two decimal places, if 0.36 g of aluminum (MM = 26.98 g/mol) are reacted. (Note: refer to the manual for the stoichiometric ratios)

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  • 1 month ago
    Best Answer

    Each mole of alum (KAl(SO₄)₂•12H₂O) contains 1 mole of Al.

    Stoichiometrically, each mole of Al can produce 1 mole of KAl(SO4)2•12H2O.

    Moles of Al reacted = (0.36 g) / (26.98 g/mol) = 0.01334 mol

    Maximum moles of KAl(SO₄)₂•12H₂O produced = 0.01334 mol

    Theoretical yield of KAl(SO₄)₂•12H₂O = (0.01334 mol) × (474.38 g/mol) = 6.33 g

    ====

    OR:

    Theoretical yield of KAl(SO₄)₂•12H₂O

    = (0.36 g Al) × (1 mol Al / 26.98 g Al) × (1 mol KAl(SO₄)₂•12H₂O / 1 mol Al) × (474.38 g KAl(SO₄)₂•12H₂O / 1 mol KAl(SO₄)₂•12H₂O)

    = 6.33 g KAl(SO₄)₂•12H₂O

  • 1 month ago

    Where is the missing data? Especially the mass of alum produced. Referring to your manual is of no value to us here.

  • david
    Lv 7
    1 month ago

    (Note: refer to the manual for the stoichiometric ratios) <<< where is that damn manual? I just keep looking for my manual, but I can't find it.

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