How many grams of lead (II) nitrate are needed to fully react with 23.5 mL of .5

Question

How many grams of lead (II) nitrate are needed to fully react with 23.5 mL of .550 M sodium chloride in the precipitation of lead (II) chlorite

Explanation / Answer

Pb(NO3)2 + 2NaCl ---> PbCl2 + 2NaNO3 So, 1 mole of lead (II) nitrate reacts with 2 moles of NaCl. Number of moles of NaCl = Molality of NaCl x Volume of the solution = 0.550 x 23.5 x 10^-3 = 12.93 x 10^-3 moles.So, number of moles of lead (II) nitrate needed = 12.93 x 10^-3 / 2 = 6.463 x 10^-3 moles. Weight of lead (II) nitrate requied = number of moles of lead (II) nitrate x molar mass = 6.463 x 10^-3 * 331.2 = 2.1404 gms

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