Answer:
Explanation:
The balanced chemical equation of this reaction;
[tex]3C_{(s)} + 2O_2{(g)} \to 2CO_{(g)}+CO_{2(g)} --- (1)[/tex]
Using the Stochiometric ratio of ratio; the product formed will be aligned with the stoichiometric ratio limiting reagent.
Thus, 3 moles of C yields 1 mole of [tex]CO_2[/tex] and 2 moles of CO.
Therefore;
1 mole of C yields 1/3 moles of [tex]CO_2[/tex] and 2/3 moles of CO
0.147 moles of C yields[tex]\dfrac{1 \times 0.159}{3}[/tex] moles of [tex]CO_2[/tex] and[tex]\dfrac{2}{3} \times 0.159[/tex] moles of CO
= 0.053 moles of [tex]CO_2[/tex] and 0.106 moles of CO
Also;
From the above reaction;
3 moles C reacts with 2 moles of [tex]O_2[/tex]
1 mole of C reacts with [tex]\dfrac{2}{3}[/tex] moles of [tex]O_2[/tex]
0.159 moles of C react with [tex]\dfrac{2}{3} \times 0.159[/tex] moles of [tex]O_2[/tex]
= 0.106 moles of [tex]O_2[/tex]
The remaining amount of [tex]O_2[/tex] = (0.117 - 0.106) mol
= 0.011 mol
The mixture comprises of the following after the reaction;
= 0.011 mol excess [tex]O_2[/tex] + 0.053 moles of [tex]CO_2[/tex] + 0.106 mol CO
= 0.17 moles of gas
Thus; moles fraction of CO is;
[tex]= \dfrac{ moles \ of \ CO \ in \ mixture }{total \ moles \ of \ gas \ in \ mixture}[/tex]
[tex]= \dfrac{0.106}{0.17}[/tex]
= 0.624
