If 1.32 g MgO is combined with 100.0 mL of 1.00 M HCl in a coffee cup calorimeter, the temperature of the HCl solution increases from 24.2 °C to 34.4 °C. Calculate the enthalpy change for the reaction per mole of MgO. Assume that the specific heat of the HCl solution is 4.18 J/g·°C and its density is 1.00 g/mL.

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

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meet024 meet024 · Answer 1 · 2022-06-29T06:43:48+02:00

The change in enthalpy for the reaction per mole is -130KJ/mol.

Given,

Reaction: MgO(s) + 2HCl (aq) →MgCl₂ (aq) + H₂O (l)

1.32 g MgO is combined with 100.0 mL of 1.00 M HCl in the cup of coffee.

Change in temperature, ΔT=34.4 °C-24.2 °C =10.2°C

Specific heat of the HCl solution, c= 4.18 J/g·°C

Density= 1.00 g/mL

Mass can be written as the product of density and volume.

Mass= Density × Volume= 1g/ml × 100 ml=100g

Heat can be written as

Heat, Q=mcΔT=100g×4.18 J/g·°C×10.2°C=4263.6J

Number of moles of the MgO,

n= (mass of MgO)/(Molar mass)= 1.32g/(40.3g/mol)= 0.03275 mol

Change in Enthalpy, ΔH=-Q/(n)=(-4263.6J)/(0.03275 mol)

ΔH= -1300186J/mol= -130KJ/mol

Hence, the change in enthalpy for the reaction per mole is -130KJ/mol.

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