Question

The rock in a particular iron ore deposit contains 82 % Fe2O3 by mass. How many kilograms of the rock must be processed to obtain 2600 kg of iron?

Answer 1

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

Answer 2

Answer:

Mass of rock = 4533.4 kg

Explanation:

Percent composition is percentage by the mass of element present in the compound.

Given that the $Fe_2O_3$ is 82 % by mass.

It means that 82 kg of $Fe_2O_3$ 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 $Fe_2O_3$

Molar mass of iron = 55.845 g/mol

Mass of 2 moles = 55.845 * 2 = 111.69 g/mol

Molar mass of $Fe_2O_3$ = 159.69 g/mol

It means,

111.69 g of iron are present in 159.69 g of $Fe_2O_3$

1 g of iron is present in 159.69/111.69 g of $Fe_2O_3$

2600*1000 g is present in (159.69/111.69)*2600*1000 g of $Fe_2O_3$

Mass of  $Fe_2O_3$ = 3717378.47 g =3717.38 kg

Thus,

82 kg of $Fe_2O_3$ is present in 100 kg of rock

1 kg of $Fe_2O_3$ is present in 100/82 kg of rock

3717.38 kg of $Fe_2O_3$ is present in (100/82)*3717.38 kg of rock

Mass of rock = 4533.4 kg