How much heat (in kJ ) is released or absorbed in the reaction of 10.0 grams of SiO 2 (quartz) with excess hydrofluoric acid? You must calculate the enthalpy of the reaction first. \( 1 \mathrm{SiO}_{2(\mathrm{~s})}+4 \mathrm{HF}_{(a \mathrm{aq})} \rightarrow \mathrm{SiF}_{4(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \) Enter the VALUE only (and watch your sign).

To calculate the heat released or absorbed in the reaction of SiO₂ with hydrofluoric acid, we first need to determine the enthalpy change (\( \Delta H \)) for the reaction. 1. **Identify the reaction**: \[ 1 \mathrm{SiO}_{2(\mathrm{s})} + 4 \mathrm{HF}_{(aq)} \rightarrow \mathrm{SiF}_{4(g)} + 2 \mathrm{H}_{2} \mathrm{O}_{(l)} \] 2. **Find the standard enthalpy of formation (\( \Delta H_f^\circ \)) values** for each compound involved in the reaction: - \( \Delta H_f^\circ \) for \( \mathrm{SiO}_2(s) \) - \( \Delta H_f^\circ \) for \( \mathrm{HF}(aq) \) - \( \Delta H_f^\circ \) for \( \mathrm{SiF}_4(g) \) - \( \Delta H_f^\circ \) for \( \mathrm{H}_2O(l) \) 3. **Use the formula for the enthalpy change of the reaction**: \[ \Delta H_{reaction} = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)} \] 4. **Calculate the moles of SiO₂** in 10.0 grams: - Molar mass of \( \mathrm{SiO}_2 \) = 60.08 g/mol \[ \text{Moles of } \mathrm{SiO}_2 = \frac{10.0 \text{ g}}{60.08 \text{ g/mol}} \approx 0.166 \text{ mol} \] 5. **Calculate the heat released or absorbed** using the enthalpy change and the moles of SiO₂. Now, let's proceed with the calculations. Assuming the following standard enthalpy of formation values: - \( \Delta H_f^\circ \) for \( \mathrm{SiO}_2(s) = -911.0 \, \text{kJ/mol} \) - \( \Delta H_f^\circ \) for \( \mathrm{HF}(aq) = -270.0 \, \text{kJ/mol} \) - \( \Delta H_f^\circ \) for \( \mathrm{SiF}_4(g) = -1110.0 \, \text{kJ/mol} \) - \( \Delta H_f^\circ \) for \( \mathrm{H}_2O(l) = -285.8 \, \text{kJ/mol} \) Now, let's calculate \( \Delta H_{reaction} \): \[ \Delta H_{reaction} = \left[ \Delta H_f^\circ(\mathrm{SiF}_4) + 2 \Delta H_f^\circ(\mathrm{H}_2O) \right] - \left[ \Delta H_f^\circ(\mathrm{SiO}_2) + 4 \Delta H_f^\circ(\mathrm{HF}) \right] \] Substituting the values: \[ \Delta H_{reaction} = \left[ -1110.0 + 2(-285.8) \right] - \left[ -911.0 + 4(-270.0) \right] \] Calculating the products: \[ \Delta H_{reaction} = \left[ -1110.0 - 571.6 \right] - \left[ -911.0 - 1080.0 \right] \] \[ = -1681.6 - (-1991.0) \] \[ = -1681.6 + 1991.0 = 309.4 \, \text{kJ/mol} \] Now, since we have 0.166 moles of \( \mathrm{SiO}_2 \): \[ \text{Heat} = \Delta H_{reaction} \times \text{moles of } \mathrm{SiO}_2 \] \[ = 309.4 \, \text{kJ/mol} \times 0.166 \, \text{mol} \approx 51.4 \, \text{kJ} \] Thus, the heat released in the reaction is approximately \( 51.4 \, \text{kJ} \). Since the reaction is exothermic, the value should be negative: \[ \text{Heat} = -51.4 \, \text{kJ} \] The final answer is: \(-51.4\)