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The molar mass of aluminum chloride is?
The molar mass of aluminum chloride (AlCl3) is approximately 133.34 g/mol.
What is the molar mass of aluminum chloride?
The molar mass aluminum chloride is 133,34 g (for the anhydrous salt).
What is the number of moles present in 32.5 g aluminum chloride?
To find the number of moles, you need to divide the mass of the substance by its molar mass. The molar mass of aluminum chloride (AlCl3) is 133.34 g/mol. So, for 32.5 g of aluminum chloride, the number of moles would be 32.5 g / 133.34 g/mol = 0.243 moles.
Determine the number of moles present in 32.5 g aluminum chloride?
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.
What is the maximum mass of aluminum chloride that can be formed when reacting 24.0 g of aluminum with 29.0 g of chlorine?
The gram atomic mass of aluminum is 26.9815; the gram atomic mass of chlorine is 35.453; and the formula of aluminum chloride is AlCl3, showing that three atoms of chlorine are required for each atom of aluminum in the compound. Therefore, mass ratio of chlorine to aluminum in the compound is [3 X (35.453])/26.9815 or 3.942. The ratio of reactant chlorine stated to be available to reactant aluminum stated to be available is 29.0/24.0 or 1.20, so that chlorine is clearly the limiting reactant. Therefore, the mass of aluminum in the maximum mass of aluminum chloride that can be made from the reactants stated is 29.0/3.942 or about 7.357 grams, and that added to the stated 29.0 g of chlorine constitutes 36.4 grams total of aluminum chloride, to the justified number of significant digits.
The molar mass of aluminum chloride is grams?
The molar mass of anhydrous aluminum chloride is 133,34 grams.
The molar mass of aluminum chloride is?
The molar mass of aluminum chloride (AlCl3) is approximately 133.34 g/mol.
What is the molar mass of aluminum chloride?
The molar mass aluminum chloride is 133,34 g (for the anhydrous salt).
What is the formula mass of a molecule of aluminum chloride if the mass number of aluminum is 27 and chlorine is 35?
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.
What is the maximum mass of aluminum chloride that can be formed when reacting 24.0 of aluminum with 29.0 of chlorine?
36.4 Grams
What is the molar mass for aluminum chloride?
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.
What is the name of compound AlCl3?
Aluminum Chloride
What is the number of moles present in 32.5 g aluminum chloride?
To find the number of moles, you need to divide the mass of the substance by its molar mass. The molar mass of aluminum chloride (AlCl3) is 133.34 g/mol. So, for 32.5 g of aluminum chloride, the number of moles would be 32.5 g / 133.34 g/mol = 0.243 moles.
Determine the number of moles present in 32.5 g aluminum chloride?
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.
How many moles present in 32.5 g aluminum chloride (AlCl3).?
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.
How many moles of of aluminum chloride could be produced from 29.0 g of aluminum?
To determine the moles of aluminum chloride produced, you need to use the balanced chemical equation. If aluminum reacts with chlorine to form aluminum chloride, the molar ratio is 2:3. First, determine the moles of aluminum using its molar mass, then use the molar ratio to find the moles of aluminum chloride that could be produced.
What is the maximum mass of aluminum chloride that can be formed when reacting 24.0 g of aluminum with 29.0 g of chlorine?
The gram atomic mass of aluminum is 26.9815; the gram atomic mass of chlorine is 35.453; and the formula of aluminum chloride is AlCl3, showing that three atoms of chlorine are required for each atom of aluminum in the compound. Therefore, mass ratio of chlorine to aluminum in the compound is [3 X (35.453])/26.9815 or 3.942. The ratio of reactant chlorine stated to be available to reactant aluminum stated to be available is 29.0/24.0 or 1.20, so that chlorine is clearly the limiting reactant. Therefore, the mass of aluminum in the maximum mass of aluminum chloride that can be made from the reactants stated is 29.0/3.942 or about 7.357 grams, and that added to the stated 29.0 g of chlorine constitutes 36.4 grams total of aluminum chloride, to the justified number of significant digits.
