An aqueous sample of 500.00mL containing Al3+ and Ca2+ was treated with an excess of 8-hydroquinone (which forms solid complexes with both metals, resulting in a precipitate containing Al(C9H6NO)3 and Ca(C9H6NO)2). The combintation of the two complexes (precipitate) weighed 1.0843g. When ignited in a furnace at 1050C, the mixture decomposed to a final form Al2O3 and CaO, weighing 0.1344g. Find the concentration of Al3+ and Ca2+ in the original sample.
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Let z be the number of moles of Al{3+} present in the original solution.
Let y be the number of moles of Ca{2+} present in the original solution.
z × (459.4341 g Al(C9H6NO)3/mol) + y × (328.3801 g Ca(C9H6NO)2/mol) = 1.0843 g
4 Al + 3 O2 → 2 Al2O3
2 Ca + O2 → 2 CaO
(z × (2/4) × (101.96137 g Al2O3/mol)) + (y × (2/2) × (56.0778 g CaO/mol)) = 0.1344 g
After some simplification these two equations remain:
459.4341 z + 328.3801 y = 1.0843
50.980685 z + 56.0778 y = 0.1344
Solve these equations simultaneously using the methods of algebra:
y = 0.00071701 and z = 0.0018476
Convert to concentrations:
(0.00071701 mol Al{3+}) / (0.50000 L) = 0.001434 mol/L Al{3+}
(0.0018476 mol Ca{2+}) / (0.50000 L) = 0.003695 mol/L Ca{2+}
Let y be the number of moles of Ca{2+} present in the original solution.
z × (459.4341 g Al(C9H6NO)3/mol) + y × (328.3801 g Ca(C9H6NO)2/mol) = 1.0843 g
4 Al + 3 O2 → 2 Al2O3
2 Ca + O2 → 2 CaO
(z × (2/4) × (101.96137 g Al2O3/mol)) + (y × (2/2) × (56.0778 g CaO/mol)) = 0.1344 g
After some simplification these two equations remain:
459.4341 z + 328.3801 y = 1.0843
50.980685 z + 56.0778 y = 0.1344
Solve these equations simultaneously using the methods of algebra:
y = 0.00071701 and z = 0.0018476
Convert to concentrations:
(0.00071701 mol Al{3+}) / (0.50000 L) = 0.001434 mol/L Al{3+}
(0.0018476 mol Ca{2+}) / (0.50000 L) = 0.003695 mol/L Ca{2+}