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?

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

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