How many grams of lead (II) chloride would be produced by reacting 50.0g of sodium chloride

=302.28 grams

Explanation:

Lead (II) Chloride is produced by the reaction between the reaction between Pb (NO₃) ₂ and NaCl according to the equation below.

Pb (NO₃) ₂ + 2NaCl → PbCl₂ + 2 NaNO₃

From the equation 2 moles of NaCl produce 1 mole of PbCl₂

Therefore the reaction ratio is 2:1

In 50.0 g of sodium, there are 50/23 moles

=2.174 moles

Since the reaction to product ratio is 2:1, then the number of moles of PbCl₂ produced is. 2.174 moles*1/2=1.087 moles.

Mass = number of moles * RMM

=1.087 moles*278.1grams/mol

=302.28 grams