When 0.726 g of carbon reacts with excess sulphur to create carbon disulphide,
C(s) + 2 S(s) → CS2(g) , the enthalpy change is 5.40 kJ. Calculate the molar enthalpy of formation of carbon disulphide.

Question

When 0.726 g of carbon reacts with excess sulphur to create carbon disulphide, 
C(s) + 2 S(s) → CS2(g) , the enthalpy change is 5.40 kJ.  Calculate the molar enthalpy of formation of carbon disulphide.

aaronq · Answer

$\Delta H_{rxn}=\Delta H^o*n$

anonymous · Answer

so we would have to take the moles of carbon disulphide and divide it by the the enthalpy change given right?

aaronq · Answer

since the limiting reactant is carbon, you'd use the moles of carbon, not carbon disulfide.

you'd end up with $\Delta H^o=\dfrac{\Delta H_{rxn}}{n_{Carbon}}$

anonymous · Answer

oh okay thanks didnt kno i had to take that into account. Thanks man got the right answer

aaronq · Answer

no problem, dude