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General Chemistry Principles, Patterns, and Applications, 2011

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A As in all other problems of this type, the first requirement is a balanced chemical equation for the reaction. Using<br />

oxidation states gives<br />

2MnO 4 − (aq) + 5HO 2 CCO 2 H(aq) + 6H + (aq) → 2Mn 2+ (aq) + 10CO 2 (g) + 8H 2 O(l)<br />

Thus each mole of MnO 4<br />

−<br />

added consumes 2.5 mol of oxalic acid.<br />

B Because we know the concentration of permanganate (0.0247 M) <strong>and</strong> the volume of permanganate solution that<br />

was needed to consume all the oxalic acid (15.4 mL), we can calculate the number of moles of MnO 4<br />

−<br />

consumed. The<br />

number of moles of oxalic acid, <strong>and</strong> thus oxalate, present can be calculated from the mole ratio of the reactants in the<br />

balanced chemical equation.<br />

C The problem asks for the percentage of calcium oxalate by mass in the original 10.0 g sample of rhubarb, so we<br />

need to know the mass of calcium oxalate that produced 9.50 × 10 −4 mol of oxalic acid. Because calcium oxalate is<br />

Ca(O 2CCO 2), 1 mol of calcium oxalate gave 1 mol of oxalic acid in the initial acid extraction:<br />

Ca(O 2 CCO 2 )(s) + 2H + (aq) → Ca 2+ (aq) + HO 2 CCO 2 H(aq)<br />

Because the problem asked for the percentage by mass of calcium oxalate in the original sample rather than for<br />

the concentration of oxalic acid in the extract, we do not need to know the volume of the oxalic acid solution for this<br />

calculation.<br />

Exercise<br />

Saylor URL: http://www.saylor.org/books<br />

Saylor.org<br />

395

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