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.
AlCl3 = 3 chloride ions per molecule and so 3 x 1.5 x 6.022 x 1023 2.7099 x 1024
Titanium (III) chloride has the formula TiCl3 and therefore contains three chloride ions per formula unit. The number of chloride ions in 0.5 mol is therefore (3/2) X Avogadro's Number or 9 X 1023, to the justified number of significant digits.
The molar mass of aluminum chloride (AlCl3) is approximately 133.34 g/mol.
The balanced chemical equation for the reaction between aluminum and chlorine to form aluminum chloride is 2Al + 3Cl2 → 2AlCl3. Using the molar masses of aluminum and chlorine, we find that 15.0 g of aluminum is equivalent to 0.56 mol and 20.0 g of chlorine is equivalent to 0.28 mol. Since aluminum and chlorine react in a 2:3 ratio, 0.56 mol of aluminum would require 0.84 mol of chlorine. Therefore, the limiting reactant is chlorine, and the maximum mass of aluminum chloride that can be formed is 59.6 g.
To determine the number of moles, you need to divide the given mass (32.5 g) by the molar mass of aluminum chloride (AlCl3). The molar mass of AlCl3 is 133.34 g/mol. So, 32.5 g / 133.34 g/mol ≈ 0.244 moles of aluminum chloride.
AlCl3 = 3 chloride ions per molecule and so 3 x 1.5 x 6.022 x 1023 2.7099 x 1024
Titanium (III) chloride has the formula TiCl3 and therefore contains three chloride ions per formula unit. The number of chloride ions in 0.5 mol is therefore (3/2) X Avogadro's Number or 9 X 1023, to the justified number of significant digits.
The molar mass of aluminum chloride (AlCl3) is approximately 133.34 g/mol.
Each compound has a different molar mass.
The molar mass of aluminum chloride (AlCl3) is approximately 133.34 g/mol. This is calculated by adding together the atomic masses of aluminum (26.98 g/mol) and chlorine (35.45 g/mol) in the compound.
The molar mass of anhydrous aluminum chloride is 133,34 grams.
To find the number of moles of aluminum chloride in 32.5 g, you first need to calculate the molar mass of AlCl3, which is 133.34 g/mol. Then, divide the given mass by the molar mass to get the number of moles. In this case, 32.5 g ÷ 133.34 g/mol ≈ 0.244 moles of aluminum chloride.
The balanced chemical equation for the reaction between aluminum and chlorine to form aluminum chloride is 2Al + 3Cl2 → 2AlCl3. Using the molar masses of aluminum and chlorine, we find that 15.0 g of aluminum is equivalent to 0.56 mol and 20.0 g of chlorine is equivalent to 0.28 mol. Since aluminum and chlorine react in a 2:3 ratio, 0.56 mol of aluminum would require 0.84 mol of chlorine. Therefore, the limiting reactant is chlorine, and the maximum mass of aluminum chloride that can be formed is 59.6 g.
To determine the number of moles, you need to divide the given mass (32.5 g) by the molar mass of aluminum chloride (AlCl3). The molar mass of AlCl3 is 133.34 g/mol. So, 32.5 g / 133.34 g/mol ≈ 0.244 moles of aluminum chloride.
The balanced chemical equation for the reaction between aluminum and chlorine is 2Al + 3Cl2 -> 2AlCl3. This means that for every 2 moles of aluminum that react, 2 moles of aluminum chloride are produced. Therefore, if 0.440 mol of aluminum is used, it will produce 0.440 mol of aluminum chloride.
you mean to say chlorine... and it 35.45g/mol
The molar mass of aluminum (Al) is 27 g/mol and chlorine (Cl) is 35 g/mol. Aluminum chloride (AlCl3) has one aluminum atom and three chlorine atoms. Therefore, the formula mass (or molar mass) of aluminum chloride is 27 (Al) + 3(35) (3Cl) = 106 g/mol.