Since I need to calculate a percent error later, would the answer for both questions be in kJ or kJ/mol and how do we find the answers?

[匿名]

2021-04-02 2:05 pm

MgO(s) + 2 HCl(aq) ⟶ MgCl2(aq) + H2O(ℓ)
given variables below
MgO: 0.52 g(mass), 40.31 g/mol(molar mass)
HCl: 3.0 mol/L(concentration), 0.050 L(volume)
specific heat capacity of water: 4.19 J/g°C
ΔT: 8.00°C

Calculate the enthalpy change, in kJ, released per mole of MgO reacted in reaction 1. Show your work.

Using Hess’ Law, calculate the theoretical enthalpy change for reaction 1 (use the values from your Chemistry 30 Data Booklet and ignore states for HCl and MgCl2).

Thank you so much for your help!

回答 (1)

Uncle Michael

2021-04-02 2:38 pm

✔ 最佳答案

MgO(s) + 2 HCl(aq) ⟶ MgCl₂(aq) + H₂O(ℓ)

Refer to the experimental data:
Initial moles of MgO = (0.52 g) / (40.31 g/mol) = 0.0129 mol
Initial moles of HCl = (3.0 mol/L) × (0.050 L) = 0.15 mol ≫ 0.0129 mol
Hence, MgO is the limiting reactant.

Heat absorbed by water = m c ΔT = [(0.050 L) × (1000 g/L)] × (4.19 J/g°C) × (8.00 °C) = 1676 J = 1.676 kJ

Enthalpy change per mole of MgO = -(1.676 kJ) / (0.0129) = -130 kJ

====
MgO(s) + 2 HCl(aq) ⟶ MgCl₂(aq) + H₂O(ℓ) ΔHrxn

Using Hess law, ΔHrxn
= ΔHf(MgCl₂) + ΔHf(H₂O) - ΔHf(MgO) - 2 ΔHf(HCl)
= (-641.3) + (-285.8) - (-601.6) - 2(-92.3) kJ
= -140.9 kJ

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