so you find the ratio of Aluminum to Oxygenin this case it is 2:3
and then because the total mole is 2.16
for aluminum:
2.16/5*2 = 0.864
for oxygen:
2.16/5*3 = 1.296
hope this helped :D
1 mole of Al2O3 has 2 moles of Al
2.16 moles Al2O3 will have 2x that number
4.32moles Al per 2.16 moles Al2O3
1 mole of Al2O3 has 2 moles of Al
2.16 moles Al2O3 will have 2x that number
4.32moles Al per 2.16 moles Al2O3
3.66x10^-3 g mol
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
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
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
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.
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.
3.66x10^-3 g mol
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
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.
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
8 mol x (4 mol / 2 mol) x 133.5 g / 1 mol = 2136 grams
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
24 mols of aluminum must have reacted since aluminum oxide is 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.
How many moles of CO2 are produced when 2.1 mol of C2H2 react?
4,51 moles hydrogen exist.
mol = mass/Mr mol = 737/ 58.44 moles of NaCl = 12.61