Question

28.__ How many moles are present in 155 g of lead (II)
acetate?

Answer 1

Write the formula

Lead (Il) acetate

Pb(C2H302)2

2) Find the molar mass of every element

Pb: 207.2 g/mol

C: 12.011 g/mol

H: 1.008 g/mol

O: 15.999 g/mol

3) Multiply each mass by the

corresponding subscript.

Pb: 207.2 g/mol* 1

207.2 g/mol

C: 12.011 g/mol* 4 =

48.044 g/mol

H: 1.008 g/mol* 6 =

6.048 g/mol

O: 15.999 g/mol* 4 =

63.996 g/mol

There are 0.4765 mol of Pb(C2H3O2)2 in 155 g