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

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Chemistry

Martha Shepard

15 March, 04:42

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

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Savannah Mcgee · 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.