If 5.00 g of sodium chloride (MM = 58.44 g/mol)  and 5.00 g of lead (II) nitrate (MM = 331.2 g/mol) are reacted according to the following balanced equation and 2.50 g of lead(II) chloride (MM = 278.1 g/mol) are obtained, what is the percent yield of the reaction? 2 NaCl(aq)  +  Pb(NO3)2(aq)  →  PbCl2(s)  +  2 NaNO3(aq)

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