Fe ions =,
Fe 2+
and
Fe 3+
Oxygen ions =,
O 2-
So, as you should see, Fe3O4, is an invalid species as the charge on the first iron ion would be 3 * 2+ = 6 +, and the charge on the second iron ion would be 3 * 3+ = 9+. This can not equal 4 * 2- = 8 -.
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 determine the amount of iron oxide (Fe2O3) formed, we first need to identify the limiting reactant. The molar mass of Fe is approximately 56 g/mol and for O2, it is about 32 g/mol. Calculating the moles, 112 g of Fe gives 2 moles (112 g / 56 g/mol), and 24 g of O2 gives 0.75 moles (24 g / 32 g/mol). The balanced reaction shows that 4 moles of Fe react with 3 moles of O2; thus, O2 is the limiting reactant. According to the stoichiometry, 0.75 moles of O2 will produce 1.0 mole of Fe2O3, which corresponds to approximately 160 g (1 mole of Fe2O3 is 160 g). Therefore, 160 g of iron oxide is formed.
The answer is 2 moles.
The answer is 10 moles of carbon monoxide.2 C + O2 = 2 CO
3 moles of O, so that must be 6 moles of O2
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 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.
First, balance the equation: 4FeS2 + 11O2 → 2Fe2O3 + 8SO2. Calculate the moles of each reactant: moles FeS2 = 1176 g / molar mass of FeS2, moles O2 = 704 g / molar mass of O2. Identify the limiting reactant based on the stoichiometry of the reaction, then use stoichiometry to calculate the grams of Fe2O3 produced.
To determine the amount of iron oxide (Fe2O3) formed, we first need to identify the limiting reactant. The molar mass of Fe is approximately 56 g/mol and for O2, it is about 32 g/mol. Calculating the moles, 112 g of Fe gives 2 moles (112 g / 56 g/mol), and 24 g of O2 gives 0.75 moles (24 g / 32 g/mol). The balanced reaction shows that 4 moles of Fe react with 3 moles of O2; thus, O2 is the limiting reactant. According to the stoichiometry, 0.75 moles of O2 will produce 1.0 mole of Fe2O3, which corresponds to approximately 160 g (1 mole of Fe2O3 is 160 g). Therefore, 160 g of iron oxide is formed.
The answer is 2 moles.
To form iron(III) oxide, the chemical equation is: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) From the equation, it can be seen that 3 moles of oxygen (O2) are needed to react with 4 moles of iron (Fe) to produce 2 moles of iron(III) oxide (Fe2O3). The molar mass of oxygen (O2) is approximately 32 g/mol. Therefore, you would need 96 grams of oxygen to react with the iron needed to form iron(III) oxide.
4
The answer is 10 moles of carbon monoxide.2 C + O2 = 2 CO
3 moles of O, so that must be 6 moles of O2
what is ag2o
15 moles O2 (32 grams/1 mole O2) = 480 grams
The answer is 0,173 moles.