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Chapter 3: Problem 82

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. If \(5.00 \mathrm{~g}\) of sulfuric acid and \(5.00 \mathrm{~g}\) of lead(II) acetate are mixed, calculate the number of grams of sulfuric acid, lead(II) acetate, lead(II) sulfate, and acetic acid present in the mixture after the reaction is complete.

Short Answer

Expert verified
After the reaction is complete, the mass of each component in the mixture is as follows: - Sulfuric acid: 0.00 g - Lead(II) acetate: 0.00 g - Lead(II) sulfate: 15.47 g - Acetic acid: 6.12 g

Step by step solution

01

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the given reaction: \(H_2SO_4 (aq) + Pb(C_2H_3O_2)_2 (aq) \rightarrow PbSO_4 (s) + 2CH_3COOH (aq)\)
02

Convert mass to moles

Next, we need to convert the mass of both reactants (sulfuric acid and lead(II) acetate) to moles. We can do this using their respective molar masses: Molar mass of \(H_2SO_4\) = 98.08 g/mol Molar mass of \(Pb(C_2H_3O_2)_2\) = 325.29 g/mol Moles of \(H_2SO_4\) = \(\frac{5.00 \thinspace g}{98.08 \thinspace g/mol} = 0.0510 \thinspace mol\) Moles of \(Pb(C_2H_3O_2)_2\) = \(\frac{5.00\thinspace g}{325.29\thinspace g/mol} = 0.01536\thinspace mol\)
03

Determine the limiting reactant

Now, we need to determine which reactant is the limiting reactant, i.e., the reactant that will be completely consumed during the reaction. In this case, we'll use the mole ratio from the balanced chemical equation: Mole ratio = \(\frac{0.0510 \thinspace mol \thinspace H_2SO_4}{1} : \frac{0.01536 \thinspace mol \thinspace Pb(C_2H_3O_2)_2}{1}\) Since the ratios are almost equal, we can tell that sulfuric acid is the limiting reactant, as there is less of it in moles.
04

Calculate the moles of products formed

Using stoichiometry, we can determine the moles of lead(II) sulfate and acetic acid formed in the reaction. Based on the balanced equation, 1 mole of sulfuric acid reacts to produce 1 mole of lead(II) sulfate and 2 moles of acetic acid: Moles of \(PbSO_4\) formed = 0.0510 mol Moles of \(CH_3COOH\) formed = 2 × 0.0510 mol = 0.102 mol
05

Calculate the moles of excess reactant remaining

Now, we will determine the amount of excess reactant (lead(II) acetate) that remains after the reaction: Moles of \(Pb(C_2H_3O_2)_2\) remaining = Initial moles - Moles reacted = 0.01536 mol - 0.0510 mol = -0.03564 mol (since negative, assume it's zero. All lead(II) acetate is used up)
06

Convert moles to mass

Finally, we will convert the moles of lead(II) sulfate, acetic acid, and unreacted lead(II) acetate back to mass using their respective molar masses: Mass of \(PbSO_4\) formed = 0.0510 mol × 303.26 g/mol = 15.47 g Mass of \(CH_3COOH\) formed = 0.102 mol × 60.05 g/mol = 6.12 g Mass of \(Pb(C_2H_3O_2)_2\) remaining = 0.000 mol × 325.29 g/mol = 0.00 g Since all the sulfuric acid was used up, its final mass in the mixture is 0.00 g. So, after the reaction, the mass of each component in the mixture is as follows: - Sulfuric acid: 0.00 g - Lead(II) acetate: 0.00 g - Lead(II) sulfate: 15.47 g - Acetic acid: 6.12 g

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Most popular questions from this chapter

Boron nitride, \(\mathrm{BN}\), is an electrical insulator with remarkable thermal and chemical stability. Its density is $2.1 \mathrm{~g} / \mathrm{cm}^{3}\(. It can be made by reacting boric acid, \)\mathrm{H}_{3} \mathrm{BO}_{3}$, with ammonia. The other product of the reaction is water. (a) Write a balanced chemical equation for the synthesis of BN. (b) If you made \(225 \mathrm{~g}\) of boric acid react with \(150 \mathrm{~g}\) ammonia, what mass of BN could you make? (c) Which reactant, if any, would be left over, and how many moles of leftover reactant would remain? (d) One application of \(\mathrm{BN}\) is as thin film for electrical insulation. If you take the mass of BN from part (a) and make a \(0.4 \mathrm{~mm}\) thin film from it, what area, in \(\mathrm{cm}^{2}\), would it cover?

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