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!