Final answer:
To react completely with 0.100 mol of Pb(NO3)2 and form a precipitate of PbCl2, 0.400 litres of 0.500M HCl are needed according to the stoichiometry which indicates two moles of HCl are required for every mole of Pb(NO3)2.
Step-by-step explanation:
To calculate how many litres of 0.500M HCl are needed to react completely with 0.100 mol of Pb(NO3)2, we first need to look at the reaction stoichiometry. The reaction shown is:
Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl₂ (s) + 2NaNO3(aq)
From this reaction, we see that one mole of Pb(NO3)2 requires two moles of NaCl to form a precipitate of PbCl2. However, we are given a reaction with HCl, and since both NaCl and HCl contain the chloride ion (Cl⁻), the stoichiometry for HCl will be the same as for NaCl. Therefore, we need two moles of HCl for every mole of Pb(NO3)2. To find the volume of 0.500M HCl needed to provide 0.100 mol of Pb(NO3)2 with enough Cl⁻ ions, we use the equation:V = n / M
Where V is volume in litres, n is the number of moles of HCl needed, and M is the molarity of the HCl solution. Since we need two moles of HCl for every mole of Pb(NO3)2:
n(HCl) = 2 moles HCl / 1 mole Pb(NO3)2 × 0.100 mol Pb(NO3)2
n(HCl) = 0.200 mol of HCl needed
V = 0.200 mol / 0.500 M
V = 0.400 L
So, we need 0.400 litres of 0.500M HCl to react completely with 0.100 mol of Pb(NO3)2 and form a precipitate of PbCl2.