Answer
$0.089\ mol$
Work Step by Step
a) $K_p=p(CO)^3p(CO_2)^3$
$p(CO)=p(CO_2)=p/2$
$K_p=(p/2)^3(p/2)^3$
$K_p=p^6/64$
$K_p=10^{-6}$
b) $K_p=p(CO)^6\rightarrow p(CO)=0.1\ atm=p(CO_2)$
$p(CO)=n_{CO}RT/V$
$n_{CO}=0.033\ mol$
$n_{Reacted}=0.01\ mol$
$n_{Unreacted}=0.100\ mol-0.01\ mol=0.089\ mol$
