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

1 Answer
anor277
Jul 29, 2017

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.

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