5.
Combustion of propane is given by :
enthalpy of combustion of propane =
"3\\times(-393.5)+ 4\\times(-285.8)-(-103.8)\\\\=2219.2\\ kJ\/ mol"
6.
Iron is the limiting reagent in this problem
Now,
224 gm of iron releases 1648 kJ energy
then, 45.8 gm of Iron will release = "\\frac{45.8}{224}\\times1648 =337 \\ kJ \\ energy"
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