Question

In the chemical reaction: 2 NaCl (s) + Pb(NO3)2 (aq) ---> 2 NaNO3 (aq) + PbCl2 (s), a student adds 2.5 grams of NaCl(s) to excess lead (II) nitrate solution. What is the theoretical mass, in grams, of the precipitate (solid) formed

Answer 1

The amount of the precipitate PbCl2 can be obtained using stoichiometry, assuming the reaction goes into completion given the excess amounts of the lead (II) nitrate solution. First, divide 2.5 g NaCl to its MW of 58.44 g/ mol to obtain the moles of NaCl involved in the reaction. Second, knowing that for every 2 moles of NaCl, there is 1 mole of PbCl2 produced, we divide the moles of NaCl obtained earlier by 2 to get the moles of PbCl2 produced. From the moles of PbCl2, we multiply it to its MW of 278.1 g/ mol. The amount of precipitate is then calculated to be 5.9484 g PbCl2.