Question

How many grams of chlorine gas must react to give 4.62 g of BiCl3?
2Bi(s)+3Cl2(g)→2BiCl3(s)

Answer 1

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!

Answer 2

biCl m(g) = 4.62g
M(g/mol) = 630.66g/mol

1) We can find M of 2BiCl3 which is 630.66g/mol

2) We can find M of 3Cl2 which is 212.718g/mol

then multiply given grams of product by ratio of Ms(2BiCl3 & 3Cl2) by which the g/mol cancels out and you get just grams of Cl2.

3) so 4.62g x 212.718g/Mol divided by 630.66g/Mol to get 1.558g of your reactant or approximately 1.6 grams depending on the significant figures the question gives you...

Good Luck!