Final answer:
The heat given off when 1.26 X10^4 g of NO_2 are produced is calculated by first determining the moles of NO_2 from its mass and then applying the heat release per mole ratio from the thermochemical equation.
Step-by-step explanation:
To determine the amount of heat given off when 1.26 \( \times \) 10^4 g of NO_2 are produced, we start by finding the number of moles of NO_2 produced using its molar mass (46.0055 g/mol). Then, we use the provided thermochemical equation 2NO(g) + O_2(g) \( \rightarrow \) 2NO_2(g) with \( \Delta H = -114.6 kJ to calculate the heat given off. This equation implies that for every 2 moles of NO_2 produced, 114.6 kJ of heat is released.
We first calculate the moles of NO_2:
Number of moles = \( \frac{1.26 \times 10^4 \text{ g}}{46.0055 \text{ g/mol}} \)
Next, we determine the heat released using the moles calculated:
Heat = (moles of NO_2) \( \times \) (\( \Delta H \) per 2 moles of NO_2)
It's important to note that since the enthalpy change is given per 2 moles, we must adjust it to the actual number of moles we calculated.