Magnetite, an iron ore with formula Fe3O4, can be reduced by treatment with hydrogen to yield iron metal and water vapor.
(a) Write the balanced equation. (Use the lowest possible coefficients.)
1Fe3O4(s) + 4H2(g)-> 3Fe(s) + 4H2O(g)
(b) This process requires 36 kcal for every 1.00 mol of Fe3O4 reduced. How much energy (in kilocalories) is required to produce 63 g of iron?
. kcal
(c) How many grams of hydrogen are needed to produce 69 g of iron?
. g
Magnetite, an iron ore with formula Fe3O4, can be reduced by treatment with hydrogen to yield iron metal and water vapor.
(a) Write the balanced equation. (Use the lowest possible coefficients.)
1Fe3O4(s) + 4H2(g)-> 3Fe(s) + 4H2O(g)
How many grams of hydrogen are needed to produce 81g of iron?
b)
Find the molecular weight of Fe3 which is 167.535
Then divide the 63g of iron by the molecular weight (167.535) and you get 0.376.
Then you multiply 0.376 by the 36 kcal to get your answer of 13.536 or rounded to 14.
36 kcal for 3 mols Fe or 12 kcal for 1 mol Fe = 55.85 g Fe.
12 kcal x (63/55.85) = ? kcal
whatis formula iron ore dencity
(a) The balanced equation is: 1Fe3O4(s) + 4H2(g) -> 3Fe(s) + 4H2O(g)
(b) To find the amount of energy required to produce 63 g of iron, we need to use the given information that 1.00 mol of Fe3O4 requires 36 kcal.
First, we need to find the molar mass of Fe3O4:
Fe3O4 = (3*55.845 g/mol) + (4*16.00 g/mol)
Fe3O4 = 231.54 g/mol
Next, we calculate the number of moles of Fe3O4 that would yield 63 g of iron:
63 g Fe / (231.54 g/mol) = 0.272 mol Fe
The amount of energy required to produce this amount of iron is:
0.272 mol Fe * 36 kcal/mol = 9.79 kcal
Therefore, it would require 9.79 kcal of energy to produce 63 g of iron.
(c) To find the number of grams of hydrogen needed to produce 69 g of iron, we can use the stoichiometry of the balanced equation. From the equation, we know that 1 mol of Fe3O4 reacts with 4 mol of H2.
First, we find the number of moles of Fe3O4 in 69 g:
69 g Fe / (231.54 g/mol) = 0.298 mol Fe
Since 1 mol of Fe3O4 requires 4 mol of H2, we can calculate the required moles of H2:
0.298 mol Fe * 4 mol H2 / 1 mol Fe3O4 = 1.192 mol H2
Finally, we convert the moles of H2 to grams:
1.192 mol H2 * (2.016 g/mol H2) = 2.40 g H2
Therefore, 2.40 g of hydrogen is needed to produce 69 g of iron.
To solve this problem, we can use stoichiometry to relate the given quantities to each other.
(a) The balanced equation is provided as:
1 Fe3O4(s) + 4 H2(g) -> 3 Fe(s) + 4 H2O(g)
(b) To determine the energy required to produce 63 g of iron, we need to use the given information that 1.00 mol of Fe3O4 requires 36 kcal. Firstly, we need to find the number of moles of iron produced from 63 g:
Molar mass of Fe = 55.845 g/mol
Number of moles of Fe = mass of Fe / molar mass of Fe
= 63 g / 55.845 g/mol
= 1.127 mol Fe
Since the molar ratio of Fe3O4:Fe is 1:3 from the balanced equation, the number of moles of Fe3O4 is 1.127 mol * (1 mol Fe3O4 / 3 mol Fe) = 0.3763 mol Fe3O4. Therefore, the energy required to produce this amount of iron is:
Energy required = 0.3763 mol Fe3O4 * 36 kcal/mol
= 13.55 kcal
Therefore, the amount of energy required to produce 63 g of iron is 13.55 kcal.
(c) To calculate the mass of hydrogen required to produce 69 g of iron, we first need to convert the mass of iron to moles:
Number of moles of Fe = mass of Fe / molar mass of Fe
= 69 g / 55.845 g/mol
= 1.235 mol Fe
Using the stoichiometric ratio from the balanced equation, the number of moles of H2 required is 4 times the number of moles of Fe:
Number of moles of H2 = 1.235 mol Fe * (4 mol H2 / 3 mol Fe)
= 1.6467 mol H2
Finally, we can find the mass of hydrogen:
Mass of H2 = number of moles of H2 * molar mass of H2
= 1.6467 mol * 2.016 g/mol
= 3.32 g H2
Therefore, 69 g of iron requires 3.32 g of hydrogen.