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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.
Calculate the number of moles in 6.80g of Al2O3?
To calculate the number of moles in a sample of Al2O3, we need to use the molar mass of Al2O3, which is 101.96 g/mol. Number of moles = Mass / Molar mass = 6.80g / 101.96 g/mol ≈ 0.067 moles. Therefore, there are approximately 0.067 moles of Al2O3 in 6.80g of the compound.
What is the number of moles of Al2O3 that are produced when 0.60 mol of Fe is produced show how you got it?
Well, honey, it's simple math. The balanced chemical equation tells us that 4 moles of Al2O3 are produced for every 3 moles of Fe. So if you start with 0.60 mol of Fe, you just set up a simple ratio and find that you'll produce 0.80 mol of Al2O3. Easy peasy lemon squeezy!
How many moles are in 6.80 g of Al2O3?
Remember the Moles Eq'n. moles = mass(g) / Mr The Mr ( Relative molecular mass of Al2O3) is . 2 x Al = 2 x 27 = 54 3 x O = 3 x 16 = 48 54 + 48 = 102 substituting moles(Al2O3) = 6.8 g / 102 moles = 0.0666.... ( recurs to infinity.
How many moles of Al203 are formed when 0.78mol o2 reacts with aluminium?
The balanced reaction equation is 4Al + 3O2 -> 2Al2O3. Therefore, 3 moles of O2 reacts with 4 moles of Al to form 2 moles of Al2O3. Since 0.78 mol of O2 is reacted, the number of moles of Al2O3 formed can be calculated using the stoichiometry of the reaction.
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.
How many moles of Al2O3 are present in 49.3 g of this compound?
To find the number of moles, we need to use the molar mass of Al2O3, which is 101.96 g/mol. Divide the given mass (49.3 g) by the molar mass to get the number of moles of Al2O3 present. 49.3 g / 101.96 g/mol ≈ 0.483 moles of Al2O3.
Calculate the number of moles in 6.80g of Al2O3?
To calculate the number of moles in a sample of Al2O3, we need to use the molar mass of Al2O3, which is 101.96 g/mol. Number of moles = Mass / Molar mass = 6.80g / 101.96 g/mol ≈ 0.067 moles. Therefore, there are approximately 0.067 moles of Al2O3 in 6.80g of the compound.
What is the mass in grams of 9.27 moles of Al2O3?
The molar mass of Al2O3 is 101.96 g/mol. To calculate the mass of 9.27 moles of Al2O3, you would multiply the moles by the molar mass: 9.27 mol x 101.96 g/mol = 945.442 g. So, the mass of 9.27 moles of Al2O3 is approximately 945.442 grams.
What is the number of moles of Al2O3 that are produced when 0.60 mol of Fe is produced show how you got it?
Well, honey, it's simple math. The balanced chemical equation tells us that 4 moles of Al2O3 are produced for every 3 moles of Fe. So if you start with 0.60 mol of Fe, you just set up a simple ratio and find that you'll produce 0.80 mol of Al2O3. Easy peasy lemon squeezy!
How many moles are in 6.80 g of Al2O3?
Remember the Moles Eq'n. moles = mass(g) / Mr The Mr ( Relative molecular mass of Al2O3) is . 2 x Al = 2 x 27 = 54 3 x O = 3 x 16 = 48 54 + 48 = 102 substituting moles(Al2O3) = 6.8 g / 102 moles = 0.0666.... ( recurs to infinity.
How many moles of Al203 are formed when 0.78mol o2 reacts with aluminium?
The balanced reaction equation is 4Al + 3O2 -> 2Al2O3. Therefore, 3 moles of O2 reacts with 4 moles of Al to form 2 moles of Al2O3. Since 0.78 mol of O2 is reacted, the number of moles of Al2O3 formed can be calculated using the stoichiometry of the reaction.
What is the theoretical yield of aluminum oxide if 2.60 mol of aluminum metal is exposed to 2.25 mol of oxygen?
the answer is 4.7 1] Figure out how many moles of Al and O2 2.5g is. 2] Compare the ratio of the moles from A to the 2:3 ratio in Al2O3; do you have more Al proportionately than O2 or vice versa ? This is called 'finding which reagent is limiting".... 3] Take whichever reagent was limiting and find out how many moles of Al2O3 you can get from it. THen find the mass.
What is the theoretical yield of aluminum oxide if 3.60 mol of aluminum metal is exposed to 3.00 mol of oxygen Express your answer with the appropriate units.?
The balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide is: 4 Al + 3 O2 -> 2 Al2O3. From the equation, 4 moles of Al react with 3 moles of O2 to produce 2 moles of Al2O3. Since there are 3.60 mol of Al (which is in excess) present and 3.00 mol of O2, 3.00 mol O2 will be the limiting reactant in this case. This will produce 2.00 mol of Al2O3. Therefore, the theoretical yield of aluminum oxide is 2.00 mol Al2O3.
What is the number of mole in 50.0g of aluminum oxide?
To find the number of moles in 50.0g of aluminum oxide, you first need to determine the molar mass of Al2O3, which is 101.96 g/mol. Then, you divide the given mass by the molar mass to get the number of moles: 50.0g / 101.96 g/mol = 0.490 moles.
Calculate the mass of alumina produced from two moles of the ore?
The concentration of alumina in the ore is necessary; also the effective yield.
How many moles of water are produced from 373 mol Al?
The balanced chemical equation for the reaction of aluminum (Al) with water (H2O) to produce aluminum oxide (Al2O3) and hydrogen gas (H2) is: 4 Al + 3 H2O → 2 Al2O3 + 3 H2. From the balanced equation, we can see that 3 moles of water are needed for every 4 moles of aluminum reacting. Therefore, if 373 mol of aluminum is reacting, then (373 mol Al) x (3 mol H2O / 4 mol Al) = 279.75 mol of water would be produced.
