Answer:
Mass of rock = 4533.4 kg
Explanation:
Percent composition is percentage by the mass of element present in the compound.
Given that the [tex]Fe_2O_3[/tex] is 82 % by mass.
It means that 82 kg of [tex]Fe_2O_3[/tex] is present in 100 kg of rock
To find the mass of rock which contains 2600 kg of iron
2600 kg = 2600*1000 g
2 moles of iron are present in 1 mole of [tex]Fe_2O_3[/tex]
Molar mass of iron = 55.845 g/mol
Mass of 2 moles = 55.845 * 2 = 111.69 g/mol
Molar mass of [tex]Fe_2O_3[/tex] = 159.69 g/mol
It means,
111.69 g of iron are present in 159.69 g of [tex]Fe_2O_3[/tex]
1 g of iron is present in 159.69/111.69 g of [tex]Fe_2O_3[/tex]
2600*1000 g is present in (159.69/111.69)*2600*1000 g of [tex]Fe_2O_3[/tex]
Mass of [tex]Fe_2O_3[/tex] = 3717378.47 g =3717.38 kg
Thus,
82 kg of [tex]Fe_2O_3[/tex] is present in 100 kg of rock
1 kg of [tex]Fe_2O_3[/tex] is present in 100/82 kg of rock
3717.38 kg of [tex]Fe_2O_3[/tex] is present in (100/82)*3717.38 kg of rock
Mass of rock = 4533.4 kg
Answer:
1625 Kg
Explanation:
Given data:
percentage of F₂O₃ = 82%
mass of iron required = 2600 g
mass of rock required = ?
Solution:
F₂O₃ contain two moles of iron.
For 2600 Kg
2600 Kg × 1 mole of F₂O₃ / 2 mole of iron = 1300 Kg
It is given that only 82% ore contain F₂O₃ .
82/100 = 0.8
1300/0.8 = 1625 Kg
