Se hace reaccionar integramente 1,2 L de acido sulfurico 0,5 M con combre metalico, de acuerdo con la reaccion: () + () + + () 2− → () 2+ + 2() + 2 Determinar la masa, en gramos, del cobre oxidado.

19.05 g have been oxidized

Explanation:

This is the unballanced equation:

H₂SO₄ + Cu → CuSO₄ + SO₂ + H₂O

Let's balance as a redox one:

Cu° oxidizes to Cu²⁺

Cu → Cu²⁺ + 2e⁻

SO₄⁻² reduces to SO₂

We add 2 water in the left side in order to ballance the O, so we add 4 protons to the right side, to ballance the H.

SO₄⁻² + 4H⁺ + 2e⁻ → SO₂ + 2H₂O

In the 2 half reactions, we have the same electrons, we sum both equations, so we have the main equation ballanced (we cancel the e⁻):

2H₂SO₄ + Cu → CuSO₄ + SO₂ + 2H₂O

We determine the moles of sulfuric acid:

M = moles /L → 0.5 mol/L . 1.2L = 0.6 moles

As ratio is 2:1, we propose

2 moles of sulfuric react to 1 mol of Cu

0.6 moles of sulfuric will react to (0.6 . 1) /2 = 0.3 moles of Cu

We determine the mass: 0.3 mol . 63.5 g/mol = 19.05 g

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explanation: