Answer:
Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL
Explanation:
Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃
For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.
Knowing 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
According to the reaction,
1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄
0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄
Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄
So, the concentration of commercial H₂SO₄ is usually 18.0M, using this as the assumed value.
Volume of H₂SO₄ = (number of H₂SO₄ required for it to be in excess)/(concentration of H₂SO₄)
Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.
QED!!!
