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How many moles og Al are in 2.16 mol of Al2O3?
so you find the ratio of Aluminum to Oxygenin this case it is 2:3and then because the total mole is 2.16for aluminum:2.16/5*2 = 0.864for oxygen:2.16/5*3 = 1.296hope this helped :D
How many moles of aluminum oxide (Al2O3) are formed in the reaction Fe2O3 plus 2Al and rarr Al2O3 plus 2Fe?
Fe2O3 + 2Al ===> Al2O3 + 2FeIn this reaction the number of moles of Al2O3 produced is dependent on the number of moles of Fe2O3 and Al that one starts with. For every 1 mole Fe2O3 and 2 moles Al, one gets 1 moles of Al2O3.
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 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 moles Al completely react?
This is a mole stoichiometry problem. Start with the balanced equation for the synthesis of aluminum oxide: 4Al + 3O2 --> 2Al2O3. The ratio of aluminum to aluminum oxide in this equation is 4:2, or 2:1, so 5.23 moles Al means half that number for Al2O3, so about 2.62 moles of aluminum oxide will be produced.
How many moles og Al are in 2.16 mol of Al2O3?
so you find the ratio of Aluminum to Oxygenin this case it is 2:3and then because the total mole is 2.16for aluminum:2.16/5*2 = 0.864for oxygen:2.16/5*3 = 1.296hope this helped :D
How many moles of aluminum oxide (Al2O3) are formed in the reaction Fe2O3 plus 2Al and rarr Al2O3 plus 2Fe?
Fe2O3 + 2Al ===> Al2O3 + 2FeIn this reaction the number of moles of Al2O3 produced is dependent on the number of moles of Fe2O3 and Al that one starts with. For every 1 mole Fe2O3 and 2 moles Al, one gets 1 moles of Al2O3.
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.
What is the percent yield of this reaction if 750.0 g of aluminum oxide produced 256.734 g of aluminum?
Aluminum oxide, Al2O3 would produce aluminum by the following decomposition:2Al2O3 ==> 4Al + 3O2 750.0 g Al2O3 x 1 mole/101.96 g = 7.356 moles Al2O3 Theoretical yield of Al = 7.356 moles Al2O3 x 4 moles Al/2 mole Al2O3 = 14.71 moles Al Theoretical mass of Al = 14.71 moles Al x 26.98 g/mole = 396.9 g Al Percent yield = 256.734 g/396.9 g (x100%) = 64.68% yield (to 4 significant figures)
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 moles Al completely react?
This is a mole stoichiometry problem. Start with the balanced equation for the synthesis of aluminum oxide: 4Al + 3O2 --> 2Al2O3. The ratio of aluminum to aluminum oxide in this equation is 4:2, or 2:1, so 5.23 moles Al means half that number for Al2O3, so about 2.62 moles of aluminum oxide will be produced.
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.
1.0Moles of AL and 1.5 Mol of oxygen what is the empirical formula?
To determine the empirical formula, first calculate the moles of each element present by dividing the given amount by their respective molar mass. For aluminum (Al), 1.0 moles of Al is equivalent to 1.0 moles, while 1.5 moles of oxygen (O) equal 1.5 moles. The ratio of Al to O is 1:1, so the empirical formula is Al2O3 (Aluminum oxide).
Calculate the number of grams of Al in 371 g of Al2O3?
To calculate the number of grams of Al in 371 g of Al2O3, you first need to determine the molar mass of Al2O3 (102 g/mol). Then, calculate the molar mass of Al (27 g/mol). From the chemical formula Al2O3, you can see that there are 2 moles of Al for every 1 mole of Al2O3. Therefore, by using these ratios, you can determine that there are 162 g of Al in 371 g of Al2O3.
How can I convert wt Al2o3 percent to Al percent?
To convert weight percent of Al2O3 to weight percent of Al, first recognize that each mole of Al2O3 contains two moles of Al. Calculate the molar mass of Al2O3 (approximately 102 g/mol) and of Al (approximately 27 g/mol). Use the formula: [ \text{wt % Al} = \left( \frac{\text{wt % Al2O3} \times 2 \times \text{molar mass of Al}}{\text{molar mass of Al2O3}} \right) ] Substituting the values gives you the weight percent of Al.
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.
How many moles of iron oxide are needed to react with 2.5 mol of Al?
The balanced chemical equation for the reaction between iron oxide (Fe2O3) and aluminum (Al) is 2Al + Fe2O3 → Al2O3 + 2Fe. This shows that 2 moles of Al react with 1 mole of Fe2O3. Therefore, 2.5 moles of Al would need 1.25 moles of Fe2O3 to completely react.
