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.
