Question

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)

Answer 1

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.

Answer 2

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.

Answer 3

He is right

Answer 4

pb(No3)2+2Nacl-72aNo3 +pbcl3

207+(14+16×3)2
207+(62)2
207+124
=331g✓

Answer 5

calculate the mass of one mole of lead ii trioxonitrate v PB(NO3)2 If pb=207, N=14, O=16

Answer 6

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)

Answer 7

U are wrong because Pb(NO3)2= MASS =(Pb=108,N=14,O=16): =[108+(14*2)+(16*6)]= (232) answer = 232.

Answer 8

Who told you lead is 108 lead is 207 and your workings are trash you are a fool ooh

Answer 9

Ur workings ar trash bt thanks to some of u

Answer 10

331g