Question

during an experiment, a student adds 1.81 g cao to 300.0 ml of 1.000 m hcl. the student observes a temperature increase of 5.00 °c.
assuming that the solutions final volume is 300.0 ml, the density is 1.00 g/ml, and the heat capacity is 4.184 j/g·°c, calculate the heat of the reaction, δh_{rxn}.
cao(s) + 2h^{+}(aq) → ca^{2+}(aq) + h_{2}o(l)

Explanation:

Step1: Calculate the mass of the solution

The volume of the solution \(V = 300.0\space mL\), and the density \(
ho=1.00\space g/mL\). Using the formula \(m=
ho V\), we have \(m = 1.00\space g/mL\times300.0\space mL=300\space g\)

Step2: Calculate the heat absorbed by the solution

The formula for heat \(q = mc\Delta T\), where \(m = 300\space g\), \(c = 4.184\space J/g\cdot^{\circ}C\), and \(\Delta T=5.00^{\circ}C\). Then \(q=(300\space g)\times(4.184\space J/g\cdot^{\circ}C)\times(5.00^{\circ}C)=6276\space J = 6.276\space kJ\)

Step3: Calculate the moles of \(CaO\)

The molar mass of \(CaO\) is \(M=(40.08 + 16.00)\space g/mol=56.08\space g/mol\). The mass of \(CaO\) is \(m_{CaO}=1.81\space g\). Using the formula \(n=\frac{m}{M}\), we get \(n_{CaO}=\frac{1.81\space g}{56.08\space g/mol}\approx0.0323\space mol\)

Step4: Calculate the heat of reaction \(\Delta H_{rxn}\)

Since the reaction is exothermic (temperature increases), \(\Delta H_{rxn}=-\frac{q}{n}\). Substituting \(q = 6.276\space kJ\) and \(n = 0.0323\space mol\), we have \(\Delta H_{rxn}=-\frac{6.276\space kJ}{0.0323\space mol}\approx - 194\space kJ/mol\)

Answer:

\(-194\space kJ/mol\)