How is the CO and N2 reaction written?

[DottZakapa]

Homework Statement

Calculate the standard enthalpy change occurring when 2.00 m^3 (measured at 0,0 C and 1 atm) of a mixture with v/v % composition of 80.0% CO and 20.0% N2 react with a stoichiometric amount of air.

Relevant Equations

thermochemistry

Can anyone explain how is this reaction ? I 've evaluated the moles of CO and N2 but now i can't see how is this reaction written
CO + N2 ...?

 

[Borek]

Hint: the only thing that reacts is CO.

 

[DottZakapa]

Hint: the only thing that reacts is CO.

$2CO + O_{2(g)} -> 2CO_{2(g)}$ correct?
then I've found the moles of CO through $n_{tot}= \frac {PV}{RT}=\frac {1atm * 2000L}{0,0821 \frac{atm*L}{mol*K}*273 K}$= 89.23 mol
Knowing that v/v%= $\frac {n_i}{n_{tot}}*100$
$n_{CO}=\frac {80,0}{100}*89,23 mol$= 71,38 mol
$\Delta H°_{f\text{ }CO_2}= -393,5 \frac{KJ}{mol}$
$\Delta H°_{f\text{ CO}}= -110,5 \frac{KJ}{mol}$

$\Delta H°_r= 2*-393,5 \frac{KJ}{mol}+2* -110,5 \frac{KJ}{mol}$=-566 $\frac{KJ}{mol}$

By proportion :
$\Delta H°_r= \frac {71,38\frac{KJ}{mol}}{2}*-566\text{ mol}$= $-20*10^3\text { KJ}$

Is all this correct?
Why in the book the result is positive? how is this possible?
Could you please explain what should i think when i see "stoichiometric amounts of air"?

 

[chemisttree]

Stoichiometric amounts of air just means that you have sufficient oxygen to completely convert all of the CO to CO2.
Book seems to be wrong as this reaction is known to be exothermic.

 

[DottZakapa]

Stoichiometric amounts of air just means that you have sufficient oxygen to completely convert all of the CO to CO2.
Book seems to be wrong as this reaction is known to be exothermic.

thanks