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. \( 1 \mathrm{SiO}_{2(\mathrm{~s})}+4 \mathrm{HF}_{(a \mathrm{aq})} \rightarrow \mathrm{SiF}_{4(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(l)} \) 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**: \[ 1 \mathrm{SiO}_{2(\mathrm{s})} + 4 \mathrm{HF}_{(aq)} \rightarrow \mathrm{SiF}_{4(\mathrm{g})} + 2 \mathrm{H}_{2} \mathrm{O}_{(l)} \] 2. **Find the enthalpy change (\( \Delta H \))**: We need the standard enthalpy of formation (\( \Delta H_f^\circ \)) values for each compound involved in the reaction. The enthalpy change for the reaction can be calculated using the formula: \[ \Delta H = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)} \] The standard enthalpy of formation values are approximately: - \( \Delta H_f^\circ \) for \( \mathrm{SiO}_2 (s) \) = -911.0 kJ/mol - \( \Delta H_f^\circ \) for \( \mathrm{HF} (aq) \) = -271.0 kJ/mol - \( \Delta H_f^\circ \) for \( \mathrm{SiF}_4 (g) \) = -1290.0 kJ/mol - \( \Delta H_f^\circ \) for \( \mathrm{H}_2O (l) \) = -285.8 kJ/mol 3. **Calculate \( \Delta H \)**: \[ \Delta H = \left[ \Delta H_f^\circ (\mathrm{SiF}_4) + 2 \Delta H_f^\circ (\mathrm{H}_2O) \right] - \left[ \Delta H_f^\circ (\mathrm{SiO}_2) + 4 \Delta H_f^\circ (\mathrm{HF}) \right] \] Plugging in the values: \[ \Delta H = \left[ -1290.0 + 2(-285.8) \right] - \left[ -911.0 + 4(-271.0) \right] \] \[ \Delta H = \left[ -1290.0 - 571.6 \right] - \left[ -911.0 - 1084.0 \right] \] \[ \Delta H = -1861.6 - (-1995.0) \] \[ \Delta H = -1861.6 + 1995.0 = 133.4 \text{ kJ/mol} \] 4. **Calculate the heat for 10.0 grams of SiO₂**: The molar mass of \( \mathrm{SiO}_2 \) is approximately \( 60.09 \text{ g/mol} \). Therefore, the number of moles in 10.0 grams is: \[ n = \frac{10.0 \text{ g}}{60.09 \text{ g/mol}} \approx 0.166 \text{ mol} \] 5. **Calculate the total heat released or absorbed**: The heat (\( q \)) can be calculated as: \[ q = n \times \Delta H \] \[ q = 0.166 \text{ mol} \times 133.4 \text{ kJ/mol} \approx 22.1 \text{ kJ} \] Since the reaction is exothermic (as indicated by the negative sign of \( \Delta H \)), the heat released will be negative. Thus, the final answer is: \[ \boxed{-22.1} \]