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What mass of Pb(NO3)2,would be required to yeild 9g of PbCl2,on d addition of excex NaCl?Pb=207,N=14,O=16,Na=23,Cl=35.5

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Pb(NO3)2 + 2NaCl>>PbCl2 + 2NaNO3(aq)
so, moles Pb(NO3)2 = moles PbCl2
Moles PbCl2= 9g/(783)
MassPb(NO3)2= molesPbCl2*(molemass leadII nitrate).

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