### Chemical Reaction AnalysisUse the chemical reaction below to answer the following questions:\[ \text{HCl(aq) + NaOH(aq) → NaCl(aq) + H}_2\text{O (l)} \hspace{10pt} \Delta H_{298}^\circ = -58 \text{ kJ/mol} \]#### Questions:(a) **How much heat is produced when 200 mL of 0.3 M HCl (density = 1.00 g/mL) and 250 mL of 0.25 M NaOH (density = 1.00 g/mL) are mixed?**(b) **If both of the solutions are at the same temperature and the heat capacity of the products is 5.39 J/g°C, how much will the temperature increase?**(c) **What assumptions are made when calculating your answers?**### Explanation:- **Reaction Details:** The reaction is an exothermic neutralization reaction, meaning it releases heat. The enthalpy change (\(\Delta H_{298}^\circ\)) is -58 kJ/mol, indicating the amount of heat released per mole of reaction.- **Calculating Heat Production:** To find out how much heat is produced, you'll need to calculate the moles of each reactant and determine the limiting reactant. Use the stoichiometric relationship to find the heat released based on the amount of limiting reactant.- **Temperature Increase:** To find how much the temperature will increase, use the formula involving specific heat capacity: \[ q = mc\Delta T \] Solve for \(\Delta T\), where \(q\) is the heat absorbed (opposite of heat released), \(m\) is the mass of the solution, and \(c\) is the specific heat capacity.- **Assumptions:** These may include ideal behavior of the solutions, complete reaction, no heat loss to the surroundings, and uniform specific heat capacity for the solution.

Given: volume of HCl (VH)=200ml volume of NaOH (VN)=250ml Molarity of HCl (MH)=0.3M Molarity of NaOH (MN)=0.25M NaOH+HCl→NaCl+H2O.......................(1)Moles of HCl=Molarity of HCl×Volume of HCl=0.3molL×200mL×L1000mL=0.06mol Moles of NaOH=Molarity of NaOH×Volume of NaOH=0.25molL×250mL×L1000mL=0.0625mol according to the reaction (1) 1 mol of HCl reacts with 1 mol of NaOH0.06mol of HCl reacts with=11×0.06 mol of NaOH =0.06mol of NaOH as moles of NaOH available are more than moles of NaOH required. Hence, HCl is limiting reagent.(a) according to the reaction (1) 1 mol of HCl produces 1 mol of NaCl0.06mol of HCl produces=11×0.06 mol of NaCl =0.06mol of NaCl ∆H°=-58kJ/mol Heat released in production of 1 mol of NaCl=58kJ Heat released in production of 0.06 mol of NaCl=0.06×58kJ=3.48kJ (b) As the initial temperature of both the solution is not given, we assume that both solution were at 298K initially. Heat capacity of product=5.39Jg°C Heat capacity of reactants=4.017Jg°C (taken from literature) as HCl is limiting reagent it is consumed completely moles of NaOH left=0.0625-0.06mol=0.0025 molmass of NaOH left=0.0025 mol×molecular weight of NaOH=0.0025 mol×40gmol=0.1g moles of NaCl produced=0.06mol mass of NaCl produced=0.06 mol×molecular weight of NaCl=0.06 mol×58.5gmol=3.51g moles of H2O produced=0.06mol mass of H2O produced=0.06 mol×molecular weight ofH2O=0.06 mol×18gmol=1.08g Heat released in reaction=heat used for increasing the temperature of products and the remaining reactant =heat used for increasing the temperature of 0.1g of NaOH+ heat used for increasing the temperature of 0.06g of NaCl+ heat used for increasing the temperature of 0.06g of H2O 3.48kJ×1000JkJ=0.1g×4.017Jg°C×T-298+0.06g×5.39Jg°C×T-298+0.06g×5.39Jg°C×T-2983480J=1.0485×T-298JT-298=34801.0485=3319T=3319+298K=3617K(c) The assumptions made for calculating the answer are: No accumulation of Heat Steady state process initial temperature of the reactants is assumed to be 298K