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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 mixture contains 3.500 g of sulfuric acid, 0 g of lead(II) acetate, 4.660 g of lead(II) sulfate, and 1.844 g of acetic acid.

Step by step solution

01

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the reaction between sulfuric acid and lead(II) acetate. This can be written as follows: \[H_2SO_4(aq) + Pb(CH_3COO)_2(aq) \rightarrow PbSO_4(s) + 2CH_3COOH(aq)\] From this equation, we see that one mole of sulfuric acid reacts with one mole of lead(II) acetate to produce one mole of lead(II) sulfate and two moles of acetic acid.
02

Find the moles of each reactant

To perform stoichiometry, we need to first determine the moles of each reactant present. We can find this using the mass of the reactant and their respective molar masses. Mass of sulfuric acid, \(H_2SO_4 = 5.00 \, g\) Molar mass of sulfuric acid, \(H_2SO_4 = 2(1.01) + 32.07 + 4(16.00) = 98.08 \, g/mol\) Moles of sulfuric acid, \(H_2SO_4 = \frac{5.00}{98.08} = 0.0510 \, mol\) Mass of lead(II) acetate, \(Pb(CH_3COO)_2 = 5.00 \, g\) Molar mass of lead(II) acetate, \(Pb(CH_3COO)_2 = 207.2 + 2(12.01 + 3(1.01) + 16.00) = 325.24 \, g/mol\) Moles of lead(II) acetate, \(Pb(CH_3COO)_2 = \frac{5.00}{325.24} = 0.01537 \, mol\)
03

Identify the limiting reactant

Now, we need to determine the limiting reactant. From the balanced chemical equation, we see that the reactants react in a 1:1 ratio. Hence, we can compare the moles of each reactant to determine which one is the limiting reactant. Since moles of sulfuric acid (0.0510 mol) are more than the moles of lead(II) acetate (0.01537 mol), the lead(II) acetate is the limiting reactant.
04

Calculate the mass of each substance present after the reaction

Now that we've identified the limiting reactant, we can determine the mass of each substance present in the mixture after the reaction. Moles of sulfuric acid consumed = moles of lead(II) acetate = 0.01537 mol Mass of sulfuric acid remaining = (0.0510 - 0.01537) mol 脳 98.08 g/mol = 3.500 g Mass of lead(II) acetate remaining = 0 g (since it's the limiting reactant) Moles of lead(II) sulfate formed = moles of lead(II) acetate = 0.01537 mol Mass of lead(II) sulfate formed = 0.01537 mol 脳 303.27 g/mol = 4.660 g Moles of acetic acid formed = 2 脳 moles of lead(II) acetate = 0.03074 mol Mass of acetic acid formed = 0.03074 mol 脳 60.05 g/mol = 1.844 g So, after the reaction is complete, the mixture contains: - 3.500 g of sulfuric acid - 0 g of lead(II) acetate - 4.660 g of lead(II) sulfate - 1.844 g of acetic acid

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Most popular questions from this chapter

Without doing any detailed calculations (hut using a periodic table to give atomic weights), rank the following samples in order of increasing numbers of atoms: \(42 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3}, 1.5 \mathrm{~mol} \mathrm{CO} 2,6.0 \times 10^{24} \mathrm{Ne}\) atoms.

A compound whose empirical formula is \(\mathrm{XF}_{3}\) consists of \(65 \%\) \(F\) by mass. What is the atomic mass of \(X\) ?

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What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{HCO}_{2}\), molar mass \(=90.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\), molar mass \(=88 \mathrm{~g} / \mathrm{mol}\)

(a) Define the terms limiting reactant and excess reactant. (b) Why are the amounts of products formed in a reaction determined only by the amount of the limiting reactant? (c) Why should you base your choice of which compound is the limiting reactant on its number of initial moles, not on its initial mass in grams?
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