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

Given:

Mass of CaO = 1.23 grams
Volume of HCl = 200 mL
Concentration of HCl = 0.500 M 
Termperature increase = 5.10 degrees celcius
Final volume = 200 mL
Density of sol'n = 1 g/mL
Cp = 4.184 J/gC

Balanced Equation:

CaO + 2HCl ---> CaCl2 + H2O

Mass of HCl = 200 mL/1000 * 0.5 mol/L * 36.46 g/mol
                      = 3.646 grams 
Mass solution = 200mL * 1 g/mL = 200 grams
H = mCpdT
    = 200 grams * 4.184 J/gC * 5.10 C
H = 4267.68 Joules 

Answer 2

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=$-194.87 \frac{KJ}{mol}$