How many grams of carbon tetrafluoride, $CF_4$ , are produced from the complete reaction of 1.46 moles of carbon in the reaction $C+2F_2 -> CF_4$ ?

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Answer 1 · 2016-07-04T06:10:42

128g

$n("CF"_4)=n("C") = 1.46mol$
$M("CF"_4)=(1xx12+4xx19)gmol^-1 = 88gmol^-1$
$m("CF"_4)=nM = 1.46molxx88gmol^-1 = 128g$ (3s.f)

How many grams of carbon tetrafluoride, CF_4CF_4, are produced from the complete reaction of 1.46 moles of carbon in the reaction C+2F_2 -> CF_4C+2F_2 -> CF_4?