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Question: 1 mole of methanol when burnt in excess of \[{{O}_{2}}\]gives out 723 KJ of heat. 1mole of \[{{O}_{2...

1 mole of methanol when burnt in excess of O2{{O}_{2}}gives out 723 KJ of heat. 1mole of O2{{O}_{2}}is used what will be the amount of heat evolved:
A. 723 KJ
B. 924 KJ
C. 482 KJ
D. 281 KJ

Explanation

Solution

Methanol is represented by molecular formula CH3OHC{{H}_{3}}OHin which alcoholic group is present due to which prefix ol is given to the compound. When methanol reacts with oxygen then the reaction proceeds is exothermic in nature i.e. a large amount of heat is produced.

Complete answer:
Reaction of the above equation can be shown as:
CH3OH+1.5O2CO2+2H2O+energyC{{H}_{3}}OH+1.5{{O}_{2}}\to C{{O}_{2}}+2{{H}_{2}}O+energy
Reaction shows that 1.5 mole of oxygen atoms are required for combustion of CH3OHC{{H}_{3}}OH
1.5 mole of O2{{O}_{2}}= 1 mole of CH3OHC{{H}_{3}}OH
Let us suppose, 1 mole O2{{O}_{2}}= x
x=11.5=0.6667x=\dfrac{1}{1.5}=0.6667mole of CH3OHC{{H}_{3}}OH
1 mole of CH3OHC{{H}_{3}}OHwith excess O2{{O}_{2}}= 723 KJ
0.667 mole of CH3OHC{{H}_{3}}OHwith excess O2{{O}_{2}}= y
y=723×0.6671=482KJy=\dfrac{723\times 0.667}{1}=482KJ

Hence we can say that 482 KJ heat is evolved i.e. option C is the correct answer.

Note:
Combustion is the process in which a substance burns in the presence of Oxygen and gives heat and light during the process. Combustion always takes place in the presence of oxygen. Combustion is classified into two types; complete combustion and incomplete combustion.