Question

How much of Ti, in g, should be obtained to have 94.97% yield?

Answer 1

Let’s us the following example.

Titanium metal can be obtained from its oxide using the following chemical reaction:

TiO₂(s) + 2C(s) → Ti(s) + 2CO₂(g)

28.6 kg of C reacts with 88.2 kg Ti(IV)O. How much of Ti, in g, should be obtained to have 94.97% yield?

M(TiO₂) = 79.86 g/mol

n(TiO₂) $= \frac{28.6 \times 10^3}{79.86} = 358.12 \;mol$

According to the reaction one mole of Ti is obtained from one mole of TiO₂.

n(Ti) =n(TiO₂) = 358.12 mol

M(Ti) = 47.90 g/mol

m(Ti) $= 47.90 \times 358.12 =17154.27 \;g = 17.154 \;kg$ (theoretical 100% yeild)

Proportion:

17.154 kg – 100 %

x kg – 94.97 %

$x = \frac{17.154 \times 94.97}{100} = 16.291 \;kg$

Answer: 16.291 kg