Question

Titanium occurs in the magnetic mineral ilmenite (FeTiO3), which is often found mixed up with sand. The ilmenite can be separated from the sand with magnets. The titanium can then be extracted from the ilmenite by the following set of reactions:
FeTiO3(s)+3Cl2(g)+3C(s)→3CO(g)+FeCl2(s)+TiCl4(g)TiCl4(g)+2Mg(s)→2MgCl2(l)+Ti(s)
Suppose that an ilmenite-sand mixture contains 20.2 % ilmenite by mass and that the first reaction is carried out with a 91.6 % yield.
If the second reaction is carried out with an 86.0 % yield, what mass of titanium can be obtained from 1.40 kg of the ilmenite-sand mixture?

Answer 1

Answer:

.222 g

Explanation:

FeTiO3(s)+3Cl2(g)+3C(s)   →    CO(g)+FeCl2(s)+TiCl4(g)

TiCl4(g)+2Mg(s)    →   2MgCl2(l)+Ti(s)

1.40 g of mineral  will contain pure ore of .202 x 1.4 g = .2828 g

ilmenite mass = .2828 g of illemnite mass

product after first reaction = .916 x .2828 = .259 g

product after 2 nd reaction = .86 x  .259 g

= .222 g

Answer 2

The mass of titanium that can be obtained from 1.40 kg of the ilmenite-sand mixture is 70.27 g

From the question,

We are to determine the mass of titanium can be obtained from 1.40 kg of the ilmenite-sand mixture

Also from the question, we have that

An ilmenite-sand mixture contains 20.2 % ilmenite by mass

That means,

The mass off ilmenite present in the 1.40 kg ilmenite-sand mixture is

= $\frac{20.2}{100} \times 1.40 \ kg$

$= \frac{20.2}{100}\times 1400 \ g$

= 20.2 × 14

= 282.8 g

∴ The mass of ilmenite present in the ilmenite-sand mixture is 282.8 g

From the first reaction

FeTiO₃(s) + 3Cl₂(g) + 3C(s) → 3CO(g) + FeCl₂(s) + TiCl₄(g)

We will calculate the mass of TiCl₄ produced

First, we will determine the number of moles of ilmenite (FeTiO₃) present

From the formula

$Number\ of \ moles = \frac{Mass }{Molar\ mass}$

Molar mass of ilmenite (FeTiO₃) = 151.71 g/mol

∴ Number of moles of ilmenite = $\frac{282.8}{151.71 }$

Number of moles of ilmenite = 1.864 moles

From the balanced chemical equation in the first reaction

1 mole of ilmenite (FeTiO₃) reacts to produce 1 mole of TiCl₄

∴ 1.864 moles of ilmenite (FeTiO₃) will produce 1.864 moles of TiCl₄

Now, for the mass of TiCl₄ produced

From the formula

Mass = Number of moles × Molar mass

Molar mass of TiCl₄ = 189.679 g/mol

∴ Mass of TiCl₄ = 1.864 × 189.679

Mass of TiCl₄= 353.561656 g

But, from the question,

The reaction has a yield of 91.6%

∴ Mass of TiCl₄ produced = $\frac{91.6}{100} \times 353.561656$

Mass of TiCl₄ produced = 0.916 × 353.561656

Mass of TiCl₄ produced = 323.8624 g

From the second reaction

TiCl₄(g) + 2Mg(s) → 2MgCl₂(l) + Ti(s)

We will determine the mass of Titanium (Ti) produced

From the balanced chemical equation for the reaction

1 mole of TiCl₄ reacts to produce 1 mole of Titanium (Ti)

The number of moles of TiCl₄ produced in the first reaction is

= 91.6% × 1.864 moles

= 0.916 × 1.864

= 1.7074242 moles

Since

1 mole of TiCl₄ reacts to produce 1 mole of Titanium (Ti)

∴ 1.707424 moles of TiCl₄ will react to produce 1.707424 moles of Titanium (Ti)

But, from the question

The yield of the second reaction is 86.0%

∴ The number of moles of Titanium (Ti) produced is $\frac{86.0}{100}\times 1.707424$

Number of moles of Ti produced = 0.86 × 1.707424

Number of moles of Ti produced  = 1.468 moles

Now, we will determine the mass of Titanium (Ti) produced

From the formula

Mass = Number of moles × Molar mass

Molar mass of Titanium = 47.867 g/mol

∴ Mass of Titanium produced = 1.468 × 47.867

Mass of Titanium produced = 70.27 g

Hence, the mass of titanium that can be obtained from 1.40 kg of the ilmenite-sand mixture is 70.27 g

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