2Pb²⁺ + O2 -----> 2Pb⁴⁺ + 2O²⁻
(Pb²⁺-2e ----> Pb⁴⁺) *2
O2 +4e ----->2O²⁻
add 2 last equations
2Pb²⁺-4e+ O2 +4e ----> 2Pb⁴⁺+ 2O²⁻
2Pb²⁺ + O2 -----> 2Pb⁴⁺ + 2O²⁻
As far as I know the Answer would be <span>True</span>
Answer:
O A. A metal higher on the activity series list will replace one that is
lower.
When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute:
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025= 0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
= 0.0075 * 63.546 =0.477 g