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

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Answer 1 · 2017-07-29T15:41:48

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

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.

What mass of C CL_4C CL_4 is formed by the reaction of 108.74 g ofmethane with 91.72 g of chlorine?