$$HCl_{(aq)}+NaOH_{(aq)}\rightarrow NaCl_{(aq)}+H_2O_{(l)}$$
The heat released by the reaction can be calculated using the following formula:
$$q=-n\Delta H$$
where q is the heat released, n is the number of moles of the limiting reactant, and \(\Delta H\) is the enthalpy change of the reaction.
To determine the limiting reactant, we need to compare the number of moles of HCl and NaOH that are present in the solution. Using the given concentrations and volumes, we can calculate the number of moles of each reactant:
$$n(HCl)=M(HCl)×V(HCl)=1.1 M×25.0 mL=27.5 ×10^{−3} mol$$
$$n(NaOH)=M(NaOH)×V(NaOH)=1.000 M×V(NaOH)$$
Since the volume of NaOH is not specified, we cannot determine the limiting reactant at this point. Let's assume that HCl is the limiting reactant and calculate the heat released by the reaction:
$$n=n(HCl)=27.5 ×10^{−3} mol$$
The enthalpy change of the reaction is \(\Delta H=-57.3 kJ/mol\). Substituting these values into the formula, we get:
$$q=-n\Delta H=-27.5 ×10^{−3} mol×(-57.3 kJ/mol)=1.57 kJ$$
Therefore, the temperature of the solution is expected to increase by 1.57 kJ.
