The answer is o,156 mol.
In the given reaction, the stoichiometry is 1:1 for Fe2O3 to Al2O3. So the number of moles of Al2O3 formed will be the same as the number of moles of Fe2O3 originally present.
To find the number of moles of Fe in 14.2 g of Fe2O3, we need to use the molar mass of Fe2O3 (molecular weight = 159.69 g/mol) and the ratio of Fe to Fe2O3. There are 2 moles of Fe in 1 mole of Fe2O3, so we find the moles of Fe in 14.2 g of Fe2O3 by: (14.2 g / 159.69 g/mol) * 2 = 0.249 moles of Fe.
To determine the number of moles of Fe2O3 in 217g of the compound, you first need to calculate the molar mass of Fe2O3, which is 159.69 g/mol. Then, divide the given mass (217g) by the molar mass to find the moles. Moles = 217g / 159.69 g/mol = 1.36 moles of Fe2O3.
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.
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.
In the given reaction, the stoichiometry is 1:1 for Fe2O3 to Al2O3. So the number of moles of Al2O3 formed will be the same as the number of moles of Fe2O3 originally present.
231 g of Fe2O3 are equal to 0,69 moles.
The balanced chemical equation for the formation of iron(III) oxide (Fe2O3) from iron (Fe) and oxygen (O2) is: 4 Fe + 3 O2 → 2 Fe2O3. From the equation, it can be seen that 3 moles of O2 are required to produce 2 moles of Fe2O3. Therefore, to produce 107.9 moles of Fe2O3, you would need (107.9 moles Fe2O3) × (3 moles O2 / 2 moles Fe2O3) = 161.85 moles of O2.
To find the number of moles of Fe in 14.2 g of Fe2O3, we need to use the molar mass of Fe2O3 (molecular weight = 159.69 g/mol) and the ratio of Fe to Fe2O3. There are 2 moles of Fe in 1 mole of Fe2O3, so we find the moles of Fe in 14.2 g of Fe2O3 by: (14.2 g / 159.69 g/mol) * 2 = 0.249 moles of Fe.
This amount may be different because rust is not a clearly definite compound.
To determine the number of moles of Fe2O3 in 217g of the compound, you first need to calculate the molar mass of Fe2O3, which is 159.69 g/mol. Then, divide the given mass (217g) by the molar mass to find the moles. Moles = 217g / 159.69 g/mol = 1.36 moles of Fe2O3.
7.18
To find the number of moles of Fe in Fe2O3, first calculate the molar mass of Fe2O3 and O. Then, determine the number of moles of O in the sample. Finally, you can use the stoichiometry of Fe2O3 to find the moles of Fe present. Alternatively, if you know the molar mass of just Fe, you can calculate the moles of Fe by dividing the mass of Fe in the sample by its molar mass.
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.
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.
Adding together the mass of two irons and three oxygen.....,251 grams Fe2O3 (1 mole Fe2O3/159.7 grams)= 1.57 moles iron II oxide ( also known as ferric oxide )===================================
To determine the number of moles of Fe that can be made from 25 moles of Fe2O3, you need to write the balanced chemical equation for producing O2 from Fe2O3. 2Fe2O3 = 4Fe + 3O2, which means that 2 moles of Fe2O3 will produce 4 moles of Fe and 3 moles of O2 . Set up a proportion. 3 moles of O2 ÷ 2 moles of Fe2O3 = x moles of O2 ÷ 25 moles of Fe2O3 Cross multiply and divide. 3 moles of O2 * 25 moles of Fe2O3 ÷ 2 moles of Fe2O3 = 37.5 moles of O2 produced.