Calculate the mass of iron that can be obtained from 1 ton of ore containing 75% red iron ore if the product yield is 80%

Red iron ore (hematite) is an ore containing iron oxide Fe2O3. The ore is often red in color.

The first step is to find out what is the mass of pure iron (III) oxide in the ore. 1 ton of ore = 1000 kg of ore.

m (Fe2O3) = 1000 * 75/100 = 750 kg of pure iron (III) oxide.

Iron is obtained by reducing it from oxide. For this, the industry uses carbon monoxide, which is formed during incomplete combustion of coke (blast furnace process). Iron recovery occurs in steps.

2 С + O2 = 2 CO

Fe2O3 + 3 CO = 2 FeO + 3 CO2

Since iron is completely reduced, its mass can be calculated through the mass fraction of iron in its oxide. The relative molecular weight of iron (III) oxide is 160, the relative atomic weight of iron is 56.

w = (2 * 56) / 160 = 0.7

Now let’s find the theoretical yield of iron from ore:

m = 750 * 0.7 = 525 kg of iron theoretical output.

Let’s calculate the practical way out:

m ‘= 525 * 80/100 = 420 kg

Answer: from 1 ton of red iron ore, the practical output of iron is 420 kg.