Question #ee172

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Answer 1 · 2017-03-03T19:49:12

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

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

And (ii) equivalent quantities of methane:

#"Moles of methane"=(438.0*g)/(16.04*g*mol^-1)=27.3*mol#.

Given the 1:1 equivalence, we could make #27.3*mol# of #"carbon tetrachloride"#, a mass of.........

#27.3*molxx153.82*g*mol^-1~=4.2*kg#. You will have to convert this into grams.