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.
Al2O3.
8 mol x (4 mol / 2 mol) x 133.5 g / 1 mol = 2136 grams
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.
Roughly 4 moles.
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.
Aluminum Oxide is Al2O3 and Al = 27 and oxygen =16 so the molar mass is 102 g/mol Al2O33.75 mol Al ~ 3.75/2 mol Al2O3 ~ (3.75/2)mol * 102 g/mol = 191.25 = 191 gram Al2O3
Al2O3.
4Fe + 3O2 -> 2Fe2O3 is the balanced equation.It'll need enough oxygen (> (0.46*3)/4 mole O2)to give (0.46*2)/4 mole Fe2O3, so 0.23 mole Fe2O3is produced.
Balanced equation: 4Al(s) + 3O2(g) ==> 2Al2O3(s)2.40 mol Al ==> 1.2 moles Al2O3 (mole ratio of Al2O3 : Al is 2:4 or 1:2)2.10 mol O2 ==> 1.4 moles Al2O3 (mole ratio of Al2O3:O2 is 2:3)Therefore, Al is the limiting reactant.Theoretical yield will thus be 1.2 moles Al2O3. If you need mass, it is 1.2 moles x 102 g/mol = 122 g
3.66x10^-3 g mol
The formula for aluminum oxide is Al2O3: Molar mass =581.77g/mol Al2O3.In one mole Al2O3, there are two moles of Al3+ ions, therefore 2 moles Al3+ x 26.982g/mol Al3+ = 53.964g Al3+ in one mole of Al2O3.GIVEN: mass of Al3+ = 53.964 g Al3+;Molar mass of Al2O3 = 581.77 g Al2O3UNKNOWN: % Al3+EQUATION:% Al3+...=...g Al3+x 100...g Al2O3% Al3+...=...53.964g Al3+x100......=...9.2758% Al3+581.77g Al2O3
35 percent of 680g = 238g
Molar Mass of Al: 2(27.0g/mol) = 54.0g/mol Molar Mass of O: 3(16.0g/mol) = 48.0g/mol Molar Mass of compound: 102.0g.mol (54.0g/mol / 102.0g/mol) x 100% = 52.9%
8 mol x (4 mol / 2 mol) x 133.5 g / 1 mol = 2136 grams
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.
143g = 5.04oz