Question #10231

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Answer 1 · 2016-07-16T05:54:06

$50%$

$"No. of moles of Oxygen" (n_(O_2))="Mass of Oxygen in gram"/"Molar mass of Oxygen"$

$=(64g)/(32g/"mol")=2" mol"$

$"No. of moles of Nitrogen"(n_(N_2))="Mass of Nitrogen in gram"/"Molar mass of Nitrogen"$

$=(14g)/(28g/"mol")=0.5" mol"$

$"No. of moles of Carbon-dioxide" (n_(CO_2))="Mass of Carbon-dioxide in gram"/"Molar mass of Carbon-dioxide"$

$=(66g)/(44g/"mol")=1.5" mol"$

So

$"mole % of Oxygen in the mixture"=(n_(O_2))/(n_(N_2)+n_(O_2)+n_(CO_2))xx100$

$=2/(2+0.5+1.5)xx100=2/4xx100=50%$