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.

ΔH=

Explanation:

The reaction between and will produce and plus energy, because is an exothermic reaction. The first step is step up the reaction:

When we balance the reaction we will get:

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):

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

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

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:

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

∆H=m*Cp*∆T

∆

Δ

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

Δ

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

Δ

ΔH=

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

the correct answer would be d. hope this !