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
2 valence electrons
Explanation:
Most transition metals have 2 valence electrons. Valence electrons are the sum total of all the electrons in the highest energy level (principal quantum number n). Most transition metals have an electron configuration that is ns2(n−1)d , so those ns2 electrons are the valence electrons.
Answer:
Explanation:
T1 = 150°C = (150 + 273.15)K = 423.15K
T2 = 45°C = (45 + 273.15)K = 318K
V1 = 693mL = 693cm³
Applying Charle's law, the volume of a given gas is directly proportional to is temperature provided that pressure remains constant.
V = kT
V1 / T1 = V2 / T2
693 / 423.15 = V2 / 318
V2 = (693 * 318) / 423.15 = 520.79cm³
The new volume of the gas is 520.79cm³
