posted by Palak on .
When 0.25M (Molarity) aqueous solutions of ammonium chloride and lead (II) nitrate are mixed a precipitate forms. How many grams of solid would be produced by mixing 250 mL of each solution?
This is a limiting reagent problem since BOTH reactants are given.
2NH4Cl + Pb(NO3)2 ==> PbCl2(s) + 2NH4NO3
1.Calculate moles NH4Cl = M x L = ??
2.Calculate moles Pb(NO3)2 = M x L = ??
3a. Using the coefficients in the balanced equation, convert moles NH4Cl to moles of the product.
3b. Same procedure, convert moles Pb(NO3)2 to moles of the product.
3c. It is quite likely, since this is a limiting reagent problem, that the answers from 3a and 3b will not agree which means one of them is wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
4. Convert moles from 3c to grams. g = moles x molar mass.