To determine if there are 143 g/mol in 6.80 g of Al₂O₃, we first need to calculate the molar mass of Al₂O₃. The molar mass of Al₂O₃ is approximately 102 g/mol (with aluminum at about 27 g/mol and oxygen at about 16 g/mol). To find the number of moles in 6.80 g of Al₂O₃, we divide the mass by the molar mass: 6.80 g ÷ 102 g/mol ≈ 0.067 moles. Thus, there are no 143 g/mol in 6.80 g of Al₂O₃; the molar mass is actually around 102 g/mol.
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.
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.
8 mol x (4 mol / 2 mol) x 133.5 g / 1 mol = 2136 grams
The concentration of alumina in the ore is necessary; also the effective yield.
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.
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!
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.
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.
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
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.
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.
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.
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.
35 percent of 680g = 238g
The gram formula mass of Al2O3 is 101.96 g/mol. This is calculated by adding the atomic masses of two aluminum atoms (26.98 g/mol) and three oxygen atoms (16.00 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
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.