Question

A mixture of0.159 moles of C is reacted with 0.117 moles of O2 in a sealed, 10.0 L-vessel at 500.0 K, producing a mixture of CO and CO2. The limiting reagent of the below reaction is carbon. For 0.159 moles of carbon, determine the amounts of products (both the CO and CO2) formed in this reaction. Also, determine the amount of O2 remaining and the mole fraction of CO when the reaction is complete.

3C(s)+2O2(g)â2CO(g)+CO2(g). 

Answer 1

Answer:

Explanation:

The balanced chemical equation of this reaction;

$3C_{(s)} + 2O_2{(g)} \to 2CO_{(g)}+CO_{2(g)} --- (1)$

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 $CO_2$  and 2 moles of CO.

Therefore;

1 mole of C yields 1/3 moles of $CO_2$  and 2/3 moles of CO

0.147 moles of C yields$\dfrac{1 \times 0.159}{3}$ moles of $CO_2$  and$\dfrac{2}{3} \times 0.159$ moles of CO

= 0.053 moles of $CO_2$  and 0.106 moles of CO

Also;

From the above reaction;

3 moles C reacts with 2 moles of $O_2$

1 mole of C reacts with $\dfrac{2}{3}$ moles of $O_2$

0.159 moles of C react with $\dfrac{2}{3} \times 0.159$ moles of $O_2$

= 0.106 moles of $O_2$

The remaining amount of $O_2$ = (0.117 - 0.106) mol

= 0.011 mol

The mixture comprises of the following after the reaction;

= 0.011 mol excess $O_2$ + 0.053 moles of $CO_2$ + 0.106 mol CO

= 0.17 moles of gas

Thus;  moles fraction of CO is;

$= \dfrac{ moles \ of \ CO \ in \ mixture }{total \ moles \ of \ gas \ in \ mixture}$

$= \dfrac{0.106}{0.17}$

= 0.624