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What is the maximum mass of aluminum chloride that can be formed when reacting 15.0 g of aluminum with 20.0 g of chlorine?
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∙ 14y ago
The maximum mass of Aluminium chloride formed will be 25.07g.
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∙ 14y ago
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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.
If 6.0g of iron reacted with 6.0g of chlorine what is the maximum number of moles of iron-II chloride that can be produced?
less than one mole
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
What is the maximum mass of aluminium chloride that can be formed when reacting 27.0 of aluminum with 32.0 of chlorine?
The answer is 40.1 g AlCl3, assuming that the questioner meant 27.0 g and 32.0 g (the units weren't provided in the question). It was calculated as follows: 27.0 g of Al is 27.0 g/27.0 g/mol = 1.00 mol of Al. 32.0 g of Cl is 32.0 g/35.45 g/mol = 0.903 mol of Cl. Aluminium chloride has the molecular formula of AlCl3, therefore it takes three moles of chlorine to form one mole of aluminum chloride, or in this case, it would take 3.00 mol of Cl to react with 1.00 mol of Al. It is clear that the amount of AlCl3 that can be formed is limited by the moles of Cl available, specifically the maximum number of moles of AlCl3 that can be formed is 0.903 mol/3 = 0.301 mol. 0.301 mol AlCl3 is equivalent to: 0.301 mol AlCl3 x 133.35 g AlCl3/mol = 40.1 g AlCl3.
What is the maximum mass of aluminum chloride that can be formed when reacting 13.0 of aluminum with 18.0 of chlorine?
I'm assuming you mean 13g of Aluminum and 18g of Chlorine Atomic mass of Al = 27g/mol number of moles = mass / atomic mass = 13g / 27g/mol = 0.481mol Molecular mass of Cl2 = 2(35.5) = 71g/mol number of moles = mass / atomic mass = 18g / 71g/mol = 0.254mol The balanced equation for reaction between Aluminum and Chlorine: 2Al + 3Cl2 --> 2AlCl3 The ratio of Al to Cl2 is 2:3 and there's only 0.254mol of Cl2 therefore it's completely reacted and Al is in excess. The ratio of Cl2 to AlCl3 is 3:2 and 0.254mol of Cl2 so there's (0.254mol x 2/3) = 0.169mol of AlCl3 formed. Molecular mass of AlCl3 : 27 + 3(35.5) = 133.5g/mol mass = number of moles x molecular mass = 0.169mol x 133.5g/mol = 22.56g of AlCl3
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 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.
What is the maximum mass of aluminum chloride that can be formed when reacting 28.0g of aluminum with 33.0g of chlorine?
The formula of aluminium chloride is AlCl3. The atomic weight of aluminium is 27 and that of chlorine is 35.5. That means 35.5*3 grams of chlorine will combine with 27 grams of aluminium. So 33 grams of chlorine will combine with 8.37 grams of aluminium. The addition of both makes it 41.37 grams. In this reaction, the whole chlorine will be utilized and only part of the aluminium.
If 6.0g of iron reacted with 6.0g of chlorine what is the maximum number of moles of iron-II chloride that can be produced?
less than one mole
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
What is the maximum mass of aluminium chloride that can be formed when reacting 27.0 of aluminum with 32.0 of chlorine?
The answer is 40.1 g AlCl3, assuming that the questioner meant 27.0 g and 32.0 g (the units weren't provided in the question). It was calculated as follows: 27.0 g of Al is 27.0 g/27.0 g/mol = 1.00 mol of Al. 32.0 g of Cl is 32.0 g/35.45 g/mol = 0.903 mol of Cl. Aluminium chloride has the molecular formula of AlCl3, therefore it takes three moles of chlorine to form one mole of aluminum chloride, or in this case, it would take 3.00 mol of Cl to react with 1.00 mol of Al. It is clear that the amount of AlCl3 that can be formed is limited by the moles of Cl available, specifically the maximum number of moles of AlCl3 that can be formed is 0.903 mol/3 = 0.301 mol. 0.301 mol AlCl3 is equivalent to: 0.301 mol AlCl3 x 133.35 g AlCl3/mol = 40.1 g AlCl3.
What is the maximum mass of aluminum chloride that can be formed when reacting 13.0 of aluminum with 18.0 of chlorine?
I'm assuming you mean 13g of Aluminum and 18g of Chlorine Atomic mass of Al = 27g/mol number of moles = mass / atomic mass = 13g / 27g/mol = 0.481mol Molecular mass of Cl2 = 2(35.5) = 71g/mol number of moles = mass / atomic mass = 18g / 71g/mol = 0.254mol The balanced equation for reaction between Aluminum and Chlorine: 2Al + 3Cl2 --> 2AlCl3 The ratio of Al to Cl2 is 2:3 and there's only 0.254mol of Cl2 therefore it's completely reacted and Al is in excess. The ratio of Cl2 to AlCl3 is 3:2 and 0.254mol of Cl2 so there's (0.254mol x 2/3) = 0.169mol of AlCl3 formed. Molecular mass of AlCl3 : 27 + 3(35.5) = 133.5g/mol mass = number of moles x molecular mass = 0.169mol x 133.5g/mol = 22.56g of AlCl3
How do you calculate the maximum mass of aluminium chloride that can be formed when reacting 15.0g of aluminium with 2.0g chlorine?
2 Al + 3Cl2 --Ã 2AlCl3 so 2 moles of Aluminium atoms react with 3 moles of chlorine molecules or 1 atom of Al reacts with 3 atoms of chlorine. Aluminium has an atomic weight of 27 and Chlorine atoms 35.5. Multiplying by Avagadro's number 1 mole of Al atoms reacts with 3 moles of chlorine atoms. So 27 g of Al reacts with 106.5 g Chlorine (3 x 35.5) to produce 133.5 g AlCl3 15g Al = 15/27 = 0.5556 moles (4 decimal places) 2g Cl = 2/35.5 = 0.0563 moles of chlorine atoms but 3 needed per molecule of AlCl3 so 0.0563/3 = 0.0188 (4 decimal places) maximum number of moles of product possible. 0.0188 x 133.5 = 2.51 g maximum yield.
When 10.0 grams of calcium reacts with 20.0 grams of chlorine gas, how many grams of calcium chloride can be produced Which reactant is in excess and which is the limiting reactant?
The balanced chemical equation for the reaction between calcium and chlorine gas to produce calcium chloride is: Ca + Cl2 -> CaCl2. From this equation, we can see that one mole of calcium reacts with one mole of chlorine gas to produce one mole of calcium chloride. The molar mass of calcium is 40.08 g/mol and the molar mass of chlorine gas is 70.90 g/mol. This means that 10.0 grams of calcium is equivalent to 0.249 moles of calcium and 20.0 grams of chlorine gas is equivalent to 0.282 moles of chlorine gas. Since the ratio of calcium to chlorine gas in the balanced chemical equation is 1:1, this means that 0.249 moles of calcium would react completely with 0.249 moles of chlorine gas, leaving an excess of 0.033 moles (or 2.34 grams) of chlorine gas. The limiting reactant in this reaction is calcium, and the maximum amount of calcium chloride that can be produced is equivalent to the number of moles of the limiting reactant, which is 0.249 moles (or 27.8 grams) of calcium chloride.
What is maximum impact spray?
Ethyl Chloride spray
What is the maximum voltage aluminum foil can conduct?
There is no maximum. Also, foil can conduct an infinite current if it is wide enough.
How does the maximum yield of products for a reaction depend on the limiting reactant?
The best way to answer this question is with an example. Using Calcium oxide reacting with hydrochloric acid, the reaction formula is: CaO + 2HCl ----->CaCl2 + H2O The molecular weight for Calcium Oxide is 56, for Hydrochloric acid is 26.5 and for calcium chloride 110. If you start with only 56g of Calcium oxide but say 10000g of hydrochloric acid, the maximum yield of the product calcium chloride can only ever be 110g. It does not matter how much hydrochloric acid is added. The limiting reactant in this example is the calcium oxide.
