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.
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.
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 find out how many moles of FeCr2O7 are required to produce 107 moles of Fe2O3, we first need to consider the balanced chemical reaction. The reaction can be represented as: [ 2 , \text{FeCr}_2\text{O}_7 \rightarrow 2 , \text{Fe}_2\text{O}_3 + 2 , \text{Cr}_2\text{O}_3 ] From the equation, 2 moles of FeCr2O7 produce 2 moles of Fe2O3, which means 1 mole of FeCr2O7 produces 1 mole of Fe2O3. Therefore, to produce 107 moles of Fe2O3, you would need 107 moles of FeCr2O7.
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.