Question

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

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

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: $$H_{2}S{O}_4(aq)+Pb(C{H}_{3}COO{)}_2(aq)\to PbS{O}_4(s)+2C{H}_{3}COOH(aq)$$ 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.

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, $H_{2}S{O}_4=5.00g$ Molar mass of sulfuric acid, $H_{2}S{O}_4=2(1.01)+32.07+4(16.00)=98.08g/mol$ Moles of sulfuric acid, $H_{2}S{O}_4=\frac{5.00}{98.08}=0.0510mol$ Mass of lead(II) acetate, $Pb(C{H}_{3}COO{)}_2=5.00g$ Molar mass of lead(II) acetate, $Pb(C{H}_{3}COO{)}_2=207.2+2(12.01+3(1.01)+16.00)=325.24g/mol$ Moles of lead(II) acetate, $Pb(C{H}_{3}COO{)}_2=\frac{5.00}{325.24}=0.01537mol$

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.

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 × 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 × 303.27 g/mol = 4.660 g Moles of acetic acid formed = 2 × moles of lead(II) acetate = 0.03074 mol Mass of acetic acid formed = 0.03074 mol × 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