find the heat of reaction deltaT = -95.7KJ
2 Al2 O3(s) > 4Al (s) + 3 O2 (g)
a. - 95.7kj
b. +95.7kJ
c. -47.8 KJ
d. +47.8KJ
e. +191.4KJ
Read and correct your post. Clarify. It makes no sense as is. delta T in kJ?
To find the heat of reaction for the given chemical equation, you need to consider the stoichiometric coefficients of the reactants and products.
In the given equation: 2 Al2 O3(s) → 4 Al (s) + 3 O2 (g)
The stoichiometric coefficient of the reactant Al2 O3 is 2, while the stoichiometric coefficient of the product Al is 4. Therefore, the heat of reaction should be multiplied by a factor of 2 to compensate for the stoichiometric difference.
Given that the heat of reaction deltaT is -95.7 KJ, the correct value to consider is:
-95.7 KJ * 2 = -191.4 KJ.
Therefore, the correct option is:
e. +191.4 KJ.