1. (5) Solid Fe reacts with aqueous sulfuric acid (H2SO4) to form aqueous iron (

Question

1. (5) Solid Fe reacts with aqueous sulfuric acid (H2SO4) to form aqueous iron (III) sulfate and hydrogen gas. Write out and balance the chemical equation, including phases.

2. (5) What type of reaction is this? Can it undergo redox? If so, write out the balanced half-reactions.

3. (10) You have 25.3 g of Fe and 65.8 g of H2SO4; which reactant is limiting for the production of hydrogen gas, and how much hydrogen gas will this produce?

4. (5) If the enthalpy for this reaction is a positive 145.61 kJ/mol, how much energy was absorbed or released during the production of the hydrogen gas above?

5. (5) If this reaction was carried out in a 6.25 L container at 21.0 °C, what is the pressure of the hydrogen gas?

Explanation / Answer

1.

2 Fe (s) + 3 H2SO4 (aq.) ----------> Fe2(SO4)3 (aq.) + 3 H2 (g)

2.

It is a hydrogen displacement reaction.

It is a redox reaction.

Oxidation half reaction :

Fe (s) ---------> Fe3+ (aq.) + 3 e-

Reduction half reaction :

2 H+ (aq.) + 2 e- ----------> H2 (g)

3.

Mass of Fe = 25.3 g.

Molar mass of Fe = 55.845g/mol

Moles of Fe = mass / molar mass = 25.3 / 55.845 = 0.453 mol

Mass of H2SO4 = 65.8 g.

Molar mass of H2SO4 = 98 g/mol

Moles of H2SO4 = 65.8 / 98.0 = 0.671 mol

From the balanced equation,

2 mol of Fe needs 3 mol of H2SO4

Then,

0.453 mol of Fe needs 3 * 0.453 / 2 = 0.680 mol of H2SO4 ( > 0.671 mol )

SO, Fe is limiting reagent.

From the balanced equation,

2 mol of Fe produces 3 mol of H2

Then, 0.453 mol of Fe produces 3 * 0.453 / 2 = 0.680 mol of H2

Mass of H2 gas produced = moles * molar mass = 0.680 * 2.0 = 1.36 g. of H2

4.

Amount heat absorbed = - 145.61 * 2 = - 291.22 kJ

5.

Ideal gas equation,

P V = n R T

P = 0.680 * 0.0821 * 294.15 / 6.25

P = 2.63 atm

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.