posted by chi on .
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)
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.
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
x = 315g x 9g divided by 278g
therefore, 10.197g of Pb(NO3)2 would be required to yield 9g of PbCl2.
calculate the mass of one mole of lead ii trioxonitrate v PB(NO3)2 If pb=207, N=14, O=16
Molar mass of Pb(NO3)2 is 207+2*14+2*3*16=331g/mol. But mole=mass/ molar mass. :- 1mole=mass (g)/331g/mol. Mass=mole*molar mass :- mass(g) = 1mole * 331g/mole =331grammes (ans)