Question

How many grams of chlorine gas must react to give 3.52g of BiCl3 according to the equation in exercise 23?

2Bi+3Cl2---->2BiCl3

Answer 1

3.52g BiCl3 × 1 mol BiCl3/ 315.34g BiCl3 × 3 mol Cl/ 2 mol BiCl3 × 70.906g Cl/ 1 mol Cl= 1.187 g Cl

Answer 2

The correct answer is 1.19 g of chlorine.

The following reaction is:

2Bi (s) + 3Cl₂ (g) ⇒ 2BiCl₃ (s)

In the reaction, it can be witnessed that 3 mol Cl₂ is equal to 2 mol BiCl₃

The molecular weight of BiCl₃ = 315.33

Thus,

3.52 g BiCl₃ = 3.52 g BiCl₃ × 1.00 mol BiCl₃ / 315.33 g BiCl₃

= 0.0112 mol BiCl₃

The mole ratio of Cl₂ and BiCl₃ is,

3 mol Cl₂ / 2 mol BiCl₃

Therefore, the amount of chlorine needed to form 0.0112 mol BiCl₃ is,

0.0112 mol BiCl₃ × 3 mol Cl₂ / 2 mol BiCl₃ = 0.0168 mol Cl₂

Now, the molecular weight of Cl₂ = 70.90

Thus,

0.0168 mol Cl₂ = 0.0168 mol Cl₂ × 70.90 g Cl₂ / 1.00 mol Cl₂

= 1.19 gm Cl₂

Hence, in the mentioned reaction, there is a need of 1.19 g of chlorine to react to produce 3.52 g of BiCl₃.