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If 5.433 g of aluminum is burned with 8.834 g of oxygen gas what is the limiting reactant?
Aluminium (Al) + Oxygen (O) = Aluminium Oxide (Al2O3)Aluminium Atomic weight = about 27Oxygen Atomic weight = about) 16Proportion in Al203 = Aluminium 54, Oxygen 48Thus the ratio of weight in Al2O3 is 54/48 = 1.125And the ratio present as reactants is 5.433/8.834 = 0.615Thus the Aluminium will run out before the Oxygen as the reaction proceeds, making the Aluminium the limiting reactant.
How many kinds of atoms in an aluminium can?
Ultrapure aluminium contain only aluminium atoms.
How many moles of aluminum oxide can be made if 5.23 mol Al completely react?
Since the ratio of moles of Al to moles of Al2O3 is 4:2, if 5.23 mol Al completely reacts, 2.615 mol Al2O3 can be made.
Given the incomplete equation 4Al plus 3O2 2X what compound is X?
The compound is Al2O3, which is aluminum oxide. This is formed when 4 moles of aluminum react with 3 moles of oxygen.
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.
How many moles of aluminum are in 10.8 grams?
Well, one mole is 26.98 g, right? Your ten grams is thus (10/26.98) moles.
How many atoms of aluminum are there in 2.89 moles of metal?
2,89 moles of aluminium contain 17,40398707673.1023 atoms.
How many grams Al from 0.4 moles AlCl3?
The mass of aluminium is 11,2 g.
How many grams of oxygen will be required to reacts completely with 0.3 moles of aluminium?
999 g
What is limiting reactant in 2Al plus 6HCl equals 2AlCl3 plus 3 H2?
2Al(s) + 6HCl(aq) = 2AlCl3(aq) + 3H2(g). The molar ratios are 2:6 :: 2:3 This means that two moles of aluminium will react to completion with exactly six moles of hydrocchloric acid. At completion there will be two moles of aluminium chloride and three moles of hydrogen gas. So if you had 8 moles of hydrochloric acid and only 2 moles of aluminium, then the aluminium would be the limiting reactant, because from the 8 moles hydrochloric acid you are using only six moles, leaving two moies on hydrochloric acid unreacted. Conversely, if you had 3 moles of Al and 6 moles HCl , then the HCl would be the limiting reactant. Because the 6 moles of acid would only react with two moles of Al leaving one mole Al unreacted.
How many moles of ions AL3 and how many mols of ions CL-has in 0.9 mols of ALCL3?
I suppose that the answers are: - 0,9 moles aluminium ions - 2,7 moles chloride ions
If 5.433 g of aluminum is burned with 8.834 g of oxygen gas what is the limiting reactant?
Aluminium (Al) + Oxygen (O) = Aluminium Oxide (Al2O3)Aluminium Atomic weight = about 27Oxygen Atomic weight = about) 16Proportion in Al203 = Aluminium 54, Oxygen 48Thus the ratio of weight in Al2O3 is 54/48 = 1.125And the ratio present as reactants is 5.433/8.834 = 0.615Thus the Aluminium will run out before the Oxygen as the reaction proceeds, making the Aluminium the limiting reactant.
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.
How many moles of aluminum ions are in 45.0 g of aluminum oxide?
Steps to take: Write the chemical formula : Al2O3 Calculate the number of moles for aluminum oxide. You should know the formula which is the mass given (0.051g) divided by the Mass of aluminum oxide. Since there is '2' moles of Aluminium you multiplly the number of moles calculated for aluminum oxide by 2. Then you have calculated the number of Ions.
How many kinds of atoms in an aluminium can?
Ultrapure aluminium contain only aluminium atoms.
How many moles of aluminum oxide can be made if 5.23 mol Al completely react?
Since the ratio of moles of Al to moles of Al2O3 is 4:2, if 5.23 mol Al completely reacts, 2.615 mol Al2O3 can be made.
Given the incomplete equation 4Al plus 3O2 2X what compound is X?
The compound is Al2O3, which is aluminum oxide. This is formed when 4 moles of aluminum react with 3 moles of oxygen.
