Answer:
Minimum volume of H₂SO₄ to ensure excess = 0.0556 mL
Explanation:
Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃
In this reaction, we consider that the specified maximum concentration of Pb(NO₃) is 0.999M, and to guarantee an excess of the other reactant, we will calculate assuming Pb(NO₃) has a concentration slightly above that, established at 1.0M.
We recognize that Concentration in mol/L = (number of moles)/(volume in L)
Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole
In accordance with the reaction,
1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄
Thus, 0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄
Consequently, the volume of H₂SO₄ needed to ensure excess is 0.001 mole of H₂SO₄
Knowing the concentration of commercial H₂SO₄ is typically 18.0M, we use this value in our calculations.
Volume of H₂SO₄ = (amount of H₂SO₄ required for excess)/(concentration of H₂SO₄)
Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.
QED!!!