1.358
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 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 find the moles of Fe3O4 needed, we first need to determine the molar ratio between Fe3O4 and Fe2O3. The balanced chemical equation for the conversion will provide this information. If the equation is Fe3O4 -> 2Fe2O3, then the molar ratio is 1:2. Thus, 4.05 moles of Fe2O3 would require 2 * 4.05 = 8.1 moles of Fe3O4.
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.
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.
231 g of Fe2O3 are equal to 0,69 moles.
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 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.
7.18
To find the moles of Fe3O4 needed, we first need to determine the molar ratio between Fe3O4 and Fe2O3. The balanced chemical equation for the conversion will provide this information. If the equation is Fe3O4 -> 2Fe2O3, then the molar ratio is 1:2. Thus, 4.05 moles of Fe2O3 would require 2 * 4.05 = 8.1 moles of Fe3O4.
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.
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.
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.
mass / molar mass molar mass Fe2O3 = 159.69 g/mol mass Fe2)3 = 4.00 kg = 4000 g moles = 4000 g / 159.69 g/mol = 25.05 moles Fe2O3 The balanced equation tells you that 1 mole Fe2O3 requires 3 moles CO to react So 25.05 moles needs (3 x 25.05) moles CO = 75.15 moles Co is needed to react 4.00 kg Fe2O3 = 75.2 mol (3 sig figs) b) The equation tells you that 1 moles Fe2O3 reacts to form 2 moles Fe So 25.05 moles will form (2 x 25.05) mol Fe moles Fe formed = 50.10 moles = 50.1 mol (3 sig figs) The equation tells you 1 mole Fe2O3 reacts to form 3 moles CO2 So 25.05 mol Fe2O3 will form (3 x 25.05) mol CO2 = 75.15 moles CO2 = 75.2 mol (3 sig figs) ==
There are two elements in the compound Fe2O3. These are: iron and oxygen