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

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.