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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.
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How many grams of aluminum chloride are produced when 18 grams of aluminum are reacted with an excess of hydrochloride acid?
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
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.
What does 6HCl 2Al 2AlCl3 3H2 mean?
This is a chemical equation describing the reaction between hydrochloric acid and aluminum to form aluminum chloride and hydrogen gas. Written with formatting, the chemical equation looks like: 6 HCl + 2 Al --> 2 AlCl3 + 3 H2
How many grams are in 5 mole of aluminum chloride?
First you take your compound AlCl3 and find the molar mass 26.98gAl+(3)35.45gCl= 133.33gAlCl3 Then you convert moles to grams 5molAlCl3x1g(over)1molx133.33g(over)1g=666.65gAlCl3 if you follow the rule of Significant Figures, 5mol has one significant figure (digit) so your answer needs to have 1 digit 700gAlCl3
Converting between moles of aluminum chloride and grams of aluminum chloride is most similar to what calculations?
Converting between moles of aluminum chloride and grams of aluminum chloride is most similar to converting between eggs and dozens of eggs. Just like one dozen equals 12 eggs, one mole of aluminum chloride contains Avogadro's number of particles (6.022 x 10^23) which corresponds to its molar mass in grams.
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 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.
Now assume you measure out exactly 2.0 ml of aluminum chloride into a third test tube How many moles of aluminum chloride are in the tube?
To calculate the number of moles of aluminum chloride in the test tube, you need to know the molar mass of aluminum chloride (AlCl3). The molar mass of AlCl3 is 133.34 g/mol. Given that you have 2.0 ml of AlCl3, you will need to convert this volume to grams using the density of AlCl3. Finally, you can calculate the moles using the molar mass.
How many grams of aluminum chloride are produced when 18 grams of aluminum are reacted with an excess of hydrochloride acid?
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
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.
How many chloride ions are present in 65.5 mL of 0.210 M AlCl3 solution?
In 1 mol of AlCl3, there are 3 chloride ions. First calculate the moles of AlCl3 in the solution: 65.5 mL is 0.0655 L. Multiply 0.0655 L by 0.210 mol/L to get the moles of AlCl3. Finally, multiply this by 3 to find the number of chloride ions in the solution.
How many moles are in 4177 g of aluminum chloride?
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.
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 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.
What is the maximum mass of aluminum chloride that can be formed when reacting 22.0 of aluminum with 27.0 of chlorine?
This equation shows how much AlCl3 can be produced with 22.0g of Al. 22.0g Al x (1 mol Al/26.98g) x (2 mol AlCl3/2mol Al) x (133.34 g/1 mol AlCl3) = 108.66g AlCl3 This equation shows how much AlCl3 can be produced with 27.0g of Cl2. 27.0g Cl2 x (1 mol Cl2/70.91g) x (2mol AlCl3/3 mol Cl2) x (133.34g/1 mol AlCl3) = 33.87g AlCl3 This shows that chlorine is the limiting reactant. Only 33.87 grams of AlCl3 can be produced before the chlorine will run out. Using significant figures, the answer is 33.8g of AlCl3
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.
What does 6HCl 2Al 2AlCl3 3H2 mean?
This is a chemical equation describing the reaction between hydrochloric acid and aluminum to form aluminum chloride and hydrogen gas. Written with formatting, the chemical equation looks like: 6 HCl + 2 Al --> 2 AlCl3 + 3 H2
