Question

During an experiment, a student adds 1.23 g of cao to 200.0 ml of 0.500 m hcl. the student observes a temperature increase of 5.10 °c. assuming the solution\'s final volume is 200.0 ml, the density if 1.00 g/ml, and the heat capacity is 4.184 j/(g·°c, calculate the heat of the reaction, ?hrxn.

Answer 1

Answer:

ΔH=$-194.87 \frac{KJ}{mol}$

Explanation:

The reaction between $HCl$ and $CaO$ will produce $H_2O$ and $CaCl_2$ plus energy, because is an exothermic reaction. The first step is step up the reaction:

$HCl + CaO -> H_2O + CaCl_2$

When we balance the reaction we will get:

$2 HCl + CaO-> H_2O + CaCl_2$

The next step is to find the moles of CaO and HCl. For the moles of CaO we need to use the molar mass of CaO (56.07 g/mol):

$1.23 g CaO \frac{1 mol CaO}{56.07gCaO} = 0.0219 g CaO$

For the moles of HCl we have to use the molarity equation (M=mol/O):

$M=\frac{mol}{L}$

$mol=M*L$

$mol=0.5 M *0.2 L=0.1 mol HCl$

If we divide by the amounts of moles in the balanced reaction is possible to find the limiting reagent:

$\frac{0.1}{1} =0.1$

$\frac{0.0219}{2} =0.01$

The limiting reagent then will be CaO. For the heat calculation process, we assume that the water on the solution is the main contributor to the total mass. Using the density value is possible to convert from mL to grams:

$200 mL\frac{1g}{1mL} =200 g$

With the mass, the heat capacity of water and the temperature increase value is possible to calculate the heat:

∆H=m*Cp*∆T

∆$H=200g*4.18\frac{J}{g^{\circ}C} *5.1^{\circ}C$

Δ$H=4267.68 J =4.26KJ$

The reaction is exothermic so, we need to chage the sign of the heat:

Δ$H=-4.26KJ$

Finally, in order to get the heat of reaction (KJ/mol) we have to divide by the moles of the limiting reagent.

Δ$H=\frac{-4.26KJ}{0.0219mol}$

ΔH=$-194.87 \frac{KJ}{mol}$

Answer 2

The heat released by the reaction is -194 KJ/mol.

The equation of the reaction is;

CaOH(aq) + 2HCl(aq) --------> CaCl2(aq) + 2H2O(l)

From the information provided in the question;

Number of moles of CaO = 1.23 g/ 56 g/mol = 0.022 moles

Number of moles of HCl =  0.500 M × 200/1000 = 0.1 moles

We now have to obtain the limiting reactant

From the reaction equation;

1 mole of CaO reacts with 2 moles of HCl

0.022 moles moles of CaO reacts with  0.022 moles × 2 moles/1 mole

= 0.044 moles

Hence there is more than enough HCl so CaO is the limiting reactant.

Temperature rise of the reaction = 5.10 °C

Total volume of solution = 200.0 ml

Density of solution =  1.00 g/ml

Mass of solution = 200 g

Heat capacity of solution = 4.184 J/g·°C

We know that, heat absorbed by solution = heat released by the reaction

Heat absorbed by solution = mcθ

m = mass of solution

c = heat capacity of solution

θ = temperature rise

Substituting values;

ΔH = 200 g × 4.184 J/g·°C × 5.10 °C

ΔH = 4.27 KJ

Therefore, heat released by the reaction;

ΔHrxn = (- 4.27 KJ)/0.022 moles

ΔHrxn = -194 KJ/mol

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