Zinc metal reacts with hydrochloric acid according to the following balanced equation.

zn(s)+2hcl(aq)→zncl2(aq)+h2(g)

when 0.121 g of zn(s) is combined with enough hcl to make 52.5 ml of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 21.8 ∘c to 23.9 ∘c. find the enthalpy change (in kj/mol) for this reaction as written. (use 1.0g/ml for the density of the solution and 4.18 j/g *celcius as the specific heat capacity.)

The enthalpy change is  -249.1 kJ/mole

Explanation:

Step 1: The balanced equation

Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)

Step 2: Data given

mass of Zn = 0.121 grams

volume of the solution = 52.5 mL = 0.0525 L

Initial temperature = 21.8°C

Final temperature = 23.9 °C

Density of the solution = 1g/mL

Specific heat capacity = 4.18 J/ g°C

Step 3: Calculating mass of solution

mass = density * volume

mass = 1g/ mL * 52.5 mL = 52.5 grams

Step 4: Calculate heat transfer

Q = m*c*ΔT

with m = mass of the solution = 52.5 grams

with c= the specific heat capacity of the solution = 4.18 J/g °C

with ΔT = the change in temperature = final - initial temperature = 23.9 - 21.8 °C = 2.1 °C

Q = 52.5 grams * 4.18 J/g°C * 2.1 °C = 460.845 J

Step 5: Calculate number of moles of Zn

moles = mass of Zinc / Molar mass of Zinc

moles of Zinc = 0.121 grams / 65.38 g/mol = 0.00185 moles

Step 6: Calculate enthalpy change

ΔHrxn = heat / moles

ΔHrxn = 460.845 J / 0.00185 moles = 249105.4 J/mole = 249.1 kJ/mole

Since reaction is exothermic, ΔHrxn is negative.

ΔHrxn = -249.1 kJ/mole

the answer is learned

it is stable as it is in group 0 which have full outer shells of electrons