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Chapter 4: Problem 144

A useful application of oxalic acid is the removal of rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) from, say, bathtub rings according to the reaction \(\begin{aligned} \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+& 6 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \longrightarrow \\ & 2 \mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}(a q)+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{H}^{+}(a q) \end{aligned}\) Calculate the number of grams of rust that can be removed by \(5.00 \times 10^{2} \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) solution of oxalic acid.

Short Answer

Expert verified
Based on the steps outlined above, first calculate the number of moles of oxalic acid in the solution. Then, determine the corresponding moles of rust using the stoichiometric relationship, and finally convert these moles of rust to grams using the molar mass of rust. The final result should be a realistic quantity of rust in grams.

Step by step solution

01

Analyze Stoichiometric Relationship

From the balanced equation, it can be see that 1 mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts with 6 moles of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\).
02

Calculate moles of oxalic acid

The given volume of the oxalic acid solution is 500.0 mL or 0.500 L and the molarity is 0.100 M. The number of moles of oxalic acid can be calculated using the formula moles = molarity x volume (in litres). So moles of oxalic acid = 0.100 M x 0.500 L = 0.0500 moles.
03

Determine moles of rust

Based on the stoichiometric relationship, 1 mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts with 6 moles of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), so 0.0500 moles of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) would correspond to \(\frac{0.0500}{6}\) moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). Calculate this fraction to get the moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\).
04

Calculate mass of rust

The molar mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is approximately 159.7 g/mol. The mass of rust that can be removed is therefore the number of moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) times this molar mass.
05

Evaluate the result

Ensure that the result makes sense in the context of the problem. The mass of rust should be a realistic quantity, and should be less than the mass of oxalic acid used, considering the stoichiometry of the reaction.

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Most popular questions from this chapter

A \(35.2-\mathrm{mL}, 1.66 M \mathrm{KMnO}_{4}\) solution is mixed with \(16.7 \mathrm{~mL}\) of \(0.892 \mathrm{M} \mathrm{KMnO}_{4}\) solution. Calculate the concentration of the final solution.

Classify the following redox reactions. (a) \(2 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (b) \(\mathrm{Mg}+2 \mathrm{AgNO}_{3} \longrightarrow \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{Ag}\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{2} \longrightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{H}_{2}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{HBr}\)

How many grams of \(\mathrm{KOH}\) are present in \(35.0 \mathrm{~mL}\) of a \(5.50 M\) solution?

(a) Describe a preparation for magnesium hydroxide \(\left[\mathrm{Mg}(\mathrm{OH})_{2}\right]\) and predict its solubility. (b) Milk of magnesia contains mostly \(\mathrm{Mg}(\mathrm{OH})_{2}\) and is effective in treating acid (mostly hydrochloric acid) indigestion. Calculate the volume of a \(0.035 \mathrm{M} \mathrm{HCl}\) solution (a typical acid concentration in an upset stomach) needed to react with two spoonfuls (approximately \(10 \mathrm{~mL}\) ) of milk of magnesia [at \(0.080 \mathrm{~g}\) \(\left.\mathrm{Mg}(\mathrm{OH})_{2} / \mathrm{mL}\right]\).

Explain how you would prepare potassium iodide (KI) by means of (a) an acid- base reaction and (b) a reaction between an acid and a carbonate compound.
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