Question

A 5 g mixture of ethanol (C₂H₅OH) and methanol (CH₃OH) reacts with excess oxygen. If this combustion releases 125.46 kJ of heat, what mass of ethanol is in the mixture? Assume that mixing does not affect any enthalpy values.

C₂H₅OH(l) + 3 O₂(g) --> 2 CO₂(g) + 3 H₂O(g)   

-277.63                              -393.5    -241.83 (kJ/mol)

CH₃OH(l) + 1.5 O₂(g) -->   CO₂(g) + 2 H₂O(g)   

-201.2                              -393.5    -241.83 (kJ/mol)

Answer 1

let x= mass of ethanol, moles of ethanol= mass/ Molecular weight, Molecular weight of ethanol= 2*12+6+16= 46

Moles of methanol= x/46, mass of me thanol =5-x, moles of methanol= (5-x)/32

from combustion of ethanol, C2H5O + 3O2-----> 2CO2 + 3H2O

enthalpy change (delH )of ethanol reaction= heat of formation of products- heat of formation of reactaants= 2*-393.5+3*-241.83 - (-277.63)= -1234.86 Kj/mol

similarly enthalpy change (delH) during combustion of methanol = -393.5+2*-241.83 +201.2= -675.34 Kh/mol

delH -ve indicates heat is released durinng the course of reaction

1 mol of ethanol gives rise to 1234.86 Kj of heat

x/46 moels give rise to x*1234.86/46 =26.84x (1)

1 mole of methanol gives rise to 675.34 K

(5-x)/32 moles gives rise to (5-x)*675.34 /32 (2)=21.1*(5-x)

Total heat generated = 125.46 Kj

heat generated due to ethanol combustion + heat generated due to methanol combustion= 125.46

26.84x (from 1)+ 21.1*(5-x)   (from 2) = 125.46

26.84x+105.5-21.1x=125.46

x*(26.84-21.1)= 125.46-105.5

x=3.47 gm mass of ethanol is mixture= 3.47 gm