What mass of iron metal can be obtained from the reaction of 45.6kg Iron (III) oxide Fe3O4 with carbon monoxide CO as given in the following equation?
Fe3O4 + 4CO = 3Fe + 4CO2
It is the same amount of iron as in the original oxide.
amountmade=percentFe*45.6kg
now the fractional percent of Fe: 3*55.8/(3*55.8+16*4)
To determine the mass of iron metal obtained from the reaction, we need to calculate the molar mass of Fe3O4 and use stoichiometry to convert from moles of Fe3O4 to moles of Fe.
Step 1: Calculate the molar mass of Fe3O4.
The molar mass of Fe3O4 can be calculated by multiplying the atomic mass of each element by the number of atoms present and summing them.
Fe3O4:
Fe: 55.85 g/mol (atomic mass of iron)
O: 16.00 g/mol (atomic mass of oxygen)
Molar mass of Fe3O4 = (3 * Fe) + (4 * O)
= (3 * 55.85) + (4 * 16.00)
= 167.55 + 64.00
= 231.55 g/mol
Step 2: Use stoichiometry to convert from moles of Fe3O4 to moles of Fe.
According to the balanced equation, the mole ratio between Fe3O4 and Fe is 3:3, which means for every 3 moles of Fe3O4, we get 3 moles of Fe.
Step 3: Convert moles of Fe to mass of Fe.
To convert moles of Fe to mass of Fe, we need to multiply the number of moles of Fe by the molar mass of Fe.
Molar mass of Fe: 55.85 g/mol (atomic mass of iron)
Now, let's calculate the mass of Fe:
1. Calculate the number of moles of Fe3O4:
Number of moles = Mass / Molar mass
= 45.6 kg / (231.55 g/mol * 1000 g/kg)
= 0.1976 mol
2. Convert moles of Fe3O4 to moles of Fe:
Moles of Fe = Moles of Fe3O4 * (3 moles of Fe / 3 moles of Fe3O4)
= 0.1976 mol * (3 / 3)
= 0.1976 mol
3. Calculate the mass of Fe:
Mass of Fe = Moles of Fe * Molar mass
= 0.1976 mol * 55.85 g/mol
= 11.03 g
Therefore, the mass of iron metal obtained from the reaction of 45.6 kg Iron (III) oxide with carbon monoxide is 11.03 g.