42.7 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 21.7 of mL of 0..415 M NaCl

Question

42.7 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 21.7 of mL of 0..415 M NaCl. The equation for the precipitate reaction is Pb(NO3)2 (aq) + 2NaCl(aq)—->PbCl 2 (s)+2 NaNO 3 (aq) The concentration of NO3^- ion in the reaction solution is what M?

Answer 1

Pb(NO₃)₂(aq) + 2NaCl(aq) ==> PbCl₂(s) + 2 NaNO₃(aq) ... balanced equation

moles Pb(NO₃)₂ = 0.0427 L x 0.255 mol/L = 0.0188 moles

moles NaCl = 0.0217 L x 0.415 mol/L = 0.00900 moles NaCl

NaCl is LIMITING

moles Pb(NO₃)₂ used up = 0.00900 mol NaCl x 1 mol Pb(NO₃)₂ / 2 mol NaCl = 0.0045 moles Pb(NO₃)₂ used

moles Pb(NO₃)₂ left = 0.0188 - 0.0045 = 0.0143 moles Pb(NO₃)₂ left

moles NO₃^- left = 0.0143 mol Pb(NO₃)₂ x 2 mol NO₃^- / mol Pb(NO₃)₂ = 0.0286 moles NO₃^- left

[NO₃^-] = 0.0286 moles / 0.0427 L + 0.0217 L = 0.0427 mol / 0.0644 L

[NO₃^-] = 0.663 M