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 ∘C to 34.1 ∘C 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∘C.

The enthalpy change per mole of Mg is (ΔH) = 460 kj mol⁻¹

Explanation:

the total volume of the solution is

100 ml, its mass is  (100 ml)(1.0 g ml⁻¹) = 100 g (Density of water 1 g ml⁻¹)

The temperature change is   ΔT = 11.1 ∘C

Heat of reaction = Cs × m × ∆T

                            = (4.18 Jg⁻¹ ∘C⁻¹)(100 g)(11.1 ∘C)

                            = 4639.8 j = 4.6 kJ

Because the process occurs at constant pressure, ΔH = qP = 4.6 kJ

To express the enthalpy change on a molar basis

Mole of Mg = = 0.01 mol

Thus, the enthalpy change per mole of Mg is ΔH = = 460 kj mol⁻¹

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b