Question

Question #f78be

Answer 1

Approx. `90*g` chlorine gas are required.

Explanation:

`Fe+3/2Cl_2 rarr FeCl_3`

`"Moles of ferric chloride"` `=` `(140*g)/(162.2*g*mol^-1)=0.863*mol.`

And given the stoichiometry we need `3/2xx2xx35.45*g*mol^-1xx0.863*mol` `=` `?*g` `*Cl_2`