2Al2O3 yields 4Al +3O2 delta H = 3351.5 kJ what is the delta H for the formation of 10.2 g of Al2 O3

Answered: 2Al2O3 yields 4Al +3O2 delta H = 3351.5…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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2Al₂O₃yields 4Al +3O₂delta H = 3351.5 kJ

what is the delta H for the formation of 10.2 g of Al₂O₃

Expert Solution

Step 1

Enthalpy of reaction is equal to the difference in enthalpy of product and reactants.

$∆ H_{reaction} = ∆ H_{product} - ∆ H_{reactant}$

Enthalpy of formation is the amount of energy required to form one mole of compounds from its elements at standard conditions.

Standard enthalpy of compound in elemental state is zero.

Number of moles is equal to the ratio of mass to molar mass.

$Number of moles = \frac{Mass}{Molar mass}$

Step 2

To determine the enthalpy of reaction.

Balanced chemical reaction

$2 {Al}_{2} O_{3} \to_{}^{} 4 Al + 3 O_{2} ∆ H = - 3351.5 kJ .$

The given enthalpy is of enthalpy of decomposition of Al₂O₃ and hence to obtained enthalpy of formation of Al₂O₃ the reaction is reversed and hence, enthalpy of formation is 3351.5kJ.

$4 Al + 3 O_{2} \to_{}^{} 2 {Al}_{2} O_{3} ∆ H = 3351.5 kJ .$

Mass of Al₂O₃=10.2g.

Molar mass of Al₂O₃=101.96g/mol.

From the balanced chemical reaction 2 mole of Al₂O₃ absorb heat 3351.5kJ.

2 mole of Al₂O₃ absorb heat= $\frac{3351.5}{2} = 1675.75 kJ .$

Number of moles of Al₂O₃= $\frac{Mass}{Molar mass}$

Number of moles of Al₂O₃= $\frac{10.2}{101.96} = 0.100 moles .$

Enthalpy of formation of Al₂O₃= $0.100 \times 1675.75 = 167.575 kJ .$

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