What mass of lead (ii) trioxonitrate(v) would be required to yield 9g of lead(ii)chloride on the addition of excess sodium chloride(pb=207,N=14,O=16,Na=23,cl=35.5)
since 315g of Pb(NO3)2 is yielded 278g of PbCl2
hence, xg of Pb(NO3)2 would yield 9g of PbCl2
it can be represented as below;
315g Pb(NO3)2 = 278g of PbCl2
xg of Pb(NO3) = 9g of PbCl2
cross multiply
x = 315g x 9g divided by 278g
x= 10.197g
therefore, 10.197g of Pb(NO3)2 would be required to yield 9g of PbCl2.
Pb(NO3)2 + 2NaCl ==> PbCl2 + 2NaNO3
mols PbCl2 = g/molar mass
Convert mols PbCl2 to mols Pb(NO3)2 using the coefficients in the balanced equation.
Now convert mols Pb(NO3)2 to g. g = mols x molar mass.
He is right
pb(No3)2+2Nacl-72aNo3 +pbcl3
207+(14+16×3)2
207+(62)2
207+124
=331g✓