How many grams of chlorine gas must react to give 4.62 g of bicl3?
2bi(s)+3cl2(g)→2bicl3(s)

Molar mass:

Cl₂ = 71.0 g/mol
BiCl₃ =  315.34 g/mol

Mole ratio:

2 Bi(s)+ 3 Cl₂(g) =2 BiCl₃(s)

3 x 71.0 g Cl₂ 2 x 315.34 g BiCl₃
?? g Cl₂ 4.62 g BiCl₃

Mass Cl₂ = 4.62 x 3 x 71.0 / 2 x 315.34 =

Mass Cl₂ = 984.06 / 630.68 = 

 1.560 g of Cl₂

hope this helps!