Aluminum chloride has a formula of AlCl3, and its molecular weight is the sum of the atomic weights of its atoms. Al has a mass of 27.0, and Cl, 35.5. So we add the masses of Al (27.0) and the chlorides (3x34.5=106.5) to get a molecular weight of 133.5 grams per mole of AlCl3.
If we have 4177g and divide it by 133.5g/mole, we get 31.30 moles.
4 moles
n = 5.670�10�3 molCl� the molar formula appears above.
That depends upon the concentration of the solution. If it is a 1 molar solution, then 2 ml contain .002 moles.
Aluminum chloride contains 3 chlorine atoms per molecular unit. Therefore, in 3 moles there are 3 times Avogadro's number of chloride ions = 1.807 X 1024.
Equation. 2Al + 3Cl2 -> 2AlCl3 one to one again 0.440 moles Al (2 moles AlCl3/2 moles Al) = 0.440 moles AlCl3 produced
1,075 moles of aluminium chloride are obtained.
4 moles
This depends on the solution concentration.
n = 5.670�10�3 molCl� the molar formula appears above.
That depends upon the concentration of the solution. If it is a 1 molar solution, then 2 ml contain .002 moles.
Aluminum chloride contains 3 chlorine atoms per molecular unit. Therefore, in 3 moles there are 3 times Avogadro's number of chloride ions = 1.807 X 1024.
Equation. 2Al + 3Cl2 -> 2AlCl3 one to one again 0.440 moles Al (2 moles AlCl3/2 moles Al) = 0.440 moles AlCl3 produced
1.26 mol of AlCl3
Four:2 Al + 3 Cl2 --> 2 AlCl3so: 4 Al + 6 Cl2 --> 4 AlCl3
1,99 grams of aluminum is equal to 0,0737 moles.
The ratio in aluminum chloride is 1:3 aluminum to chloride ( AlCl3 or Al2Cl6 )
10 grams aluminum (1 mole Al/26.98 grams) = 0.37 moles of aluminum ---------------------------------