0
What else can I help you with?
How many moles of chloride ions are in 0.2520g of aluminum chloride?
To find the number of moles of chloride ions in aluminum chloride, you first need to convert 0.2520g of aluminum chloride to moles. Then, since there are three chloride ions per one aluminum chloride molecule, you would multiply the number of moles of aluminum chloride by 3 to find the moles of chloride ions.
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.
When 4 moles of aluminum are allowed to react with an excess of chlorine gas Cl2 how many moles of aluminum chloride are produced?
When 4 moles of aluminum react with an excess of chlorine gas, 4 moles of aluminum chloride are produced. This is because the balanced chemical equation for the reaction is: 2Al + 3Cl2 -> 2AlCl3 This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride, so 4 moles of aluminum will produce 4 moles of aluminum chloride.
How many chloride ions are there in 3 mol of aluminum chloride?
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.
You are given 34.0 g of aluminum and 39.0 g of chlorine gas. If you had excess chlorine how many moles of aluminum chloride could be produced from 34.0 g of aluminum?
First, calculate the number of moles of aluminum in 34.0 g using its molar mass. Then, determine the limiting reactant by converting the moles of aluminum to moles of aluminum chloride using the mole ratio from the balanced chemical equation. Finally, calculate the moles of aluminum chloride that can be produced based on the limiting reactant.
How many moles of chloride ions are in 0.2520g of aluminum chloride?
To find the number of moles of chloride ions in aluminum chloride, you first need to convert 0.2520g of aluminum chloride to moles. Then, since there are three chloride ions per one aluminum chloride molecule, you would multiply the number of moles of aluminum chloride by 3 to find the moles of chloride ions.
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.
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.
When 4 moles of aluminum are allowed to react with an excess of chlorine gas Cl2 how many moles of aluminum chloride are produced?
When 4 moles of aluminum react with an excess of chlorine gas, 4 moles of aluminum chloride are produced. This is because the balanced chemical equation for the reaction is: 2Al + 3Cl2 -> 2AlCl3 This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride, so 4 moles of aluminum will produce 4 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.
How many chloride ions are there in 3 mol of aluminum chloride?
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.
You are given 34.0 g of aluminum and 39.0 g of chlorine gas. If you had excess chlorine how many moles of aluminum chloride could be produced from 34.0 g of aluminum?
First, calculate the number of moles of aluminum in 34.0 g using its molar mass. Then, determine the limiting reactant by converting the moles of aluminum to moles of aluminum chloride using the mole ratio from the balanced chemical equation. Finally, calculate the moles of aluminum chloride that can be produced based on the limiting reactant.
When 4 moles of aluminum are allowed to react with an excess of chlorine gas CL2 how many moles of aluminum chlorde are produced?
When 4 moles of aluminum react with an excess of chlorine gas (Cl2), 4 moles of aluminum chloride are produced because the balanced chemical equation for this reaction is: 2 Al + 3 Cl2 -> 2 AlCl3 Since the mole ratio between aluminum and aluminum chloride is 2:2, it means that for every 2 moles of aluminum, 2 moles of aluminum chloride are produced.
How many moles of aluminum are present in an aluminum cylinder with a mass of 15 g?
To determine the number of moles of aluminum present, we need to first determine the molar mass of aluminum, which is approximately 26.98 g/mol. We can then use the formula: moles = mass / molar mass. Plugging in the values, we get moles = 15 g / 26.98 g/mol ≈ 0.56 moles of aluminum.
If you had excess aluminum how many moles of aluminum chloride could be produced from 31.0 g of chlorine gas Cl2?
To calculate the moles of aluminum chloride produced, you would first need to determine the limiting reactant. Compare the moles of each reactant (Aluminum and Cl2) using their molar masses. Whichever reactant produces fewer moles of aluminum chloride would be the limiting reactant. Once you have that, you can use the stoichiometry of the balanced chemical equation to calculate the moles of aluminum chloride produced.
How many moles of sulfur are present in 3 moles of aluminum sulfate?
There are 6 moles of sulfur present in 3 moles of aluminum sulfate, because aluminum sulfate has a 2:3 ratio of aluminum to sulfur.
When 0.440 mol of aluminum are allowed to react with an excess of chlorine gas Cl2 how many moles of aluminum chloride are produced?
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.
