When 0.243 g of mg metal is combined with enough hcl to make 100 ml of solution in a constant-pressure calorimeter, the following reaction occurs: mg(s) + 2 hcl(aq) → mgcl2(aq) + h2(g) if the temperature of the solution increases from 23.0 ºc to 34.1 ºc as a result of this reaction, calculate δh in kj/mol of mg. assume that the solution has a specific heat of 4.18 j/g−ºc.

ΔH = - 464 kJ/mol  

Explanation:

Step 1: Data given

Mass of Mg = 0.243 grams

volume of solution = 100 mL

Initial temperature = 23.0 °C

Final temperature = 34.1 °C

Specific heat = 4.18 J/g°C

Step 2: Calculate the heat transfer

Q= m*c*ΔT

⇒ m = the mass of the solution = 100 grams

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

⇒ ΔT = The change in temperature = T2 - T1 = 34.1 - 23.0 = 11.1 °C

Q = 100 g * 4.18 J/g°C * 11.1°C

Q = 4639.8 J

Step 3: Calculate moles of Mg

Moles Mg = 0.243 grams / 24.3 g/mol

Moles Mg = 0.01 moles

Step 4: Calculate  ΔH

ΔH = 4639.8 J / 0.01 moles

ΔH = 463980 J/mol = 464 kJ/mol

(4.6398 kJ) / (0.0099979 mol) = 464 kJ/mol  

Since the reaction is exothermic, the ΔH is negative by convention, so:

ΔH = - 464 kJ/mol  

answer; i believe the correct answer is (solid gold);

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