How many grams of lead (2) chloride can be collected from 34.5 g sodium chloride and an excess amount of lead (2) nitrate?

119.9 grams of Lead chloride will be collected.

Explanation:

PbNO3 + 2NaCl PbCl2+2NaNO3

From the equation we know that 2 moles of NaCl is required to convert 1 mole of PbNO3 to PbCl2.

Lead nitrate here, is a limiting agent:

No of moles of NaCl used is calculated as the mass is given, atomic mass of NaCl is 40

No. of moles= wt/atomic weight

                       =  34.5/40

                        = 0.8625 moles of Nacl will be required.

So from stoichiometry,

1/2=  x/0.862      

0.862=2x

x=0.862/2

 = 0.431 moles

So x is moles of PbCl2 precipitated

From the formula

n=wt/at wt

wt= n*at wt

    = 0.431*278.1  ( 278.1 is the atomic weight of lead chloride)

     = 119.9 grams of PbCl2

lead nitrate required will be 1 mole of Lead nitrate amounts to 331.2 gms.

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