Question

What mass of `C CL_4` is formed by the reaction of 108.74 g ofmethane with 91.72 g of chlorine?

Answer 1

Approx. `38*g` `"CCl"_4`......

Explanation:

We need (i) a stoichiometric equation......

`"CH"_4(g) + "4Cl"_2(g) rarr"CCl"_4+"4HCl(g)"`

And (ii) equivalent quantities of methane and dichlorine,

`"Moles of methane"` `=` `(108.74*g)/(16.04*g*mol^-1)=6.78*mol`

`"Moles of dichlorine"` `=` `(91.72*g)/(70.90*g*mol^-1)=1.29*mol`

Clearly chlorine gas is the limiting reagent. And at most we can make `1.29/4*mol` `"carbon tet"....=(153.81*g*mol^-1)/4=38.5*g` perhalide.