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.
