Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 5.80 g of sulfuric acid and 5.80 g of lead(II) acetate are mixed.

part a-d

Answer 1

Part A: After the reaction, approximately 7.08g of acetic acid is present.

Part B: After the reaction, approximately 5.46g of lead(II) sulfate is present.

Part C: After the reaction, the same amount of lead(II) sulfate (5.46g) is present.

Part D: After the reaction, approximately 7.08g of acetic acid is present.

Part A: Calculating the grams of sulfuric acid present after the reaction.

The balanced chemical equation for the reaction is:

`H_(2)SO_(4) + Pb(CH_(3)COO)_(2) (aq)` →
`PbSO_(4)(s) + 2CH_(3)COOH(aq)`

The molar mass of sulfuric acid (
`H_(2) SO_(4)`)98.08g/mol. Using the given mass of sulfuric acid (5.80g), we can calculate the moles of sulfuric acid:

Moles of
`H_(2) SO_(4)` =
`(Mass)/(Molar Mass )` =
`(5.80g)/(98.08g/mol )`

Moles of
`H_(2) SO_(4)` ≈0.059mol

After the reaction, every mole of sulfuric acid produces one mole of acetic acid. So, the moles of acetic acid formed (2 × Moles of 2×Moles of
`H_(2) SO_(4)` ):

Moles of acetic acid =2×0.059 mol

Now, we can find the mass of acetic acid
`(CH_(3) COOH)` using its molar mass (60.05g/mol):

Mass of acetic acid = Moles of acetic acid × Molar Mass of acetic acid

Mass of acetic acid=2×0.059mol×60.05g/mol

Mass of acetic acid ≈ 7.08g

Part B: Calculating the grams of lead(II) acetate present after the reaction.

The molar mass of lead(II) acetate (
`Pb(CH_(3) COO) _(2)`) is 325.29g/mol. Using the given mass of lead(II) acetate (5.80g), we can calculate the moles of lead(II) acetate:

Mass Molar Mass = 5.80 g 325.29 g/mol

Moles of
`Pb(CH_(3) COO) _(2)`​ =
`(Mass)/(Molar Mass )` =
`(5.80g)/(325.29g/mol )`

Moles of
`Pb(CH_(3) COO) _(2)` ≈0.018mol

After the reaction, every mole of lead(II) acetate produces one mole of lead(II) sulfate(
`PbSO_(4)`). So, the moles of lead(II) sulfate formed:

Moles of
`PbSO_(4)` =0.018mol

Now, we can find the mass of lead(II) sulfate using its molar mass
`(303.26g/mol)`:

Mass of
`PbSO_(4)` =Moles of
`PbSO_(4)` ×Molar Mass of
`PbSO_(4)`

​Mass of
`PbSO_(4)` =0.018mol×303.26g/mol

Mass of
`PbSO_(4)` ≈5.46g

Part C: Calculating the grams of lead(II) sulfate present after the reaction.

The moles of lead(II) sulfate formed (0.018mol) are the same as calculated in Part B.

Part D: Calculating the grams of acetic acid present after the reaction.

The mass of acetic acid (7.08g) is the same as calculated in Part A.