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tester [92]
3 years ago
13

2 Al + 3 H2SO4 --> Al2(SO4)3 + 3 H2

Chemistry
1 answer:
suter [353]3 years ago
3 0

Answer:

The answer to your question is 0.15 g of H₂

Explanation:

Data

mass of H₂ =

mass of H₂SO₄ = 7.5 g

Balanced chemical reaction

             2Al  +  3H₃SO₄  ⇒   Al₂(SO₄)₃  +  3H₂

Process

1.- Calculate the molar mass of hydrogen and sulfuric acid

H₂ = 3(2) = 6 g

H₂SO₄ = 2 + 32 + (16 x 4) = 3 x 98 g = 294 g

2.- Use proportions to solve this problem

            294 g of H₂SO₄ ---------------- 6 g of H₂

              7.5 g of H₂SO₄ ----------------  x

                        x = (7.5 x 6)/294

                        x = 45/294

                        x = 0.15 g of H₂

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Nitric oxide reacts with chlorine to form nocl. the data refer to 298 k. 2no(g) + cl2(g) → 2nocl(g) substance: no(g) cl2(g) nocl
tigry1 [53]

Answer:

- 10.555 kJ/mol.

Explanation:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

T is the temperature of the reaction (K).

∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

  • Calculating ∆H°rxn:

∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

  • Calculating ∆S°rxn:

∵  ∆S°rxn = ∑∆S°products - ∑∆S°reactants

<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>

<em></em>

  • Calculating ∆G°rxn:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>

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