Answer:
C
Explanation:
We can use heat stoichiometry. Thus, convert from grams of Ba to moles of Ba; and use the ΔH value to calculate the amount of heat released. Hence:
[tex]\displaystyle 5.75\text{ g Ba}\cdot \frac{1\text{ mol Ba}}{137.33\text{ g Ba}} \cdot \frac{1\text{ mol}_{rxn}}{2\text{ mol Ba}} \cdot \frac{-1107\text{ kJ}}{1\text{ mol}_{rxn}} =-46.4 \text{ kJ}[/tex]
Therefore, about 46.4 kJ of heat was released.
In conclusion, our answer is C.