If you are given 718 grams of lead (iv) oxalate, how many atoms of oxygen do you have?

Answer 1

Pb(C2O4)2

718g of Pb(C2O4)2 *(1 mole of Pb(C2O4)2/383.2g)= moles of Pb(C2O4)2

moles of Pb(C2O4)2 *(8 moles of O/1 mole of Pb(C2O4)2)= moles of O

Moles of O*(6.02 x 10^23 atoms/1 mole of O)= atoms of O