Consider the following reaction: Mn(s) + CuSO4(aq) → MnSO4(aq) + Cu(s) Which of the following statements regarding this reaction is correct?

The sulfate ion is oxidized.

Each copper gains two electrons.

Manganese is the oxidizing agent.

Manganese is neither oxidized nor reduced.

Copper is the reducing agent.
Please answer quickly

Each copper gains two electrons

Explanation:

Oxidizing agents:

Oxidizing agents oxidize the other elements and itself gets reduced.

Reducing agents:

Reducing agents reduced the other element are it self gets oxidized.

Chemical equation:

CuSO₄ + Mn   →  MnSO₄  + Cu

This is single replacement reaction. Manganese replace the copper and react with sulfate to form copper sulfate.

Copper gains two electrons and gets reduced. The oxidation state of copper on left side is +2 while on right side its 0 so it gain two electrons and gets reduced.

while oxidation state of manganese on left side is 0 while on right side its +2. It means Mn lose to electrons and gets oxidized.

So  copper is oxidizing agent while manganese is reducing agent.

while the oxidation state of sulfate is not changed.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

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