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Trava [24]
3 years ago
9

A beaker contains nickel(ii sulfate, zinc(ii sulfate and solid iron. what species in this beaker will be reduced?

Chemistry
2 answers:
Serggg [28]3 years ago
6 0
If a beaker is contained with the substances nickel sulfate, zinc sulfate and solid iron, there is a substance that will be reduced in the following substances that is in it. The substance that will be reduced as it is in the beaker is the Ni^2 + or also known as the Nickel (II)
attashe74 [19]3 years ago
4 0

Answer:

Nickel (II) to solid nickel.

Explanation:

Hello,

In this case, the placed species are nickel (II) sulfate NiSO_4, zinc sulfate ZnSO_4 and solid iron Fe. Now, as the nickel sulfate is more reactive than the zinc sulfate, it will experience the following reduction half-reaction from nickel (II) to solid nickel:

Ni^{2+}+2e^--->Ni^0

Nevertheless, it is possible that the zinc might experience the same reduction reaction from zinc (II) to solid zinc as well.

Best regards.

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Answer:

310.53 g of Cu.

Explanation:

The balanced equation for the reaction is given below:

CuSO₄ + Zn —> ZnSO₄ + Cu

Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:

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Molar mass of Cu = 63.5 g/mol

Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g

Summary:

From the balanced equation above,

159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.

Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:

From the balanced equation above,

159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.

Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.

Thus, 310.53 g of Cu were obtained from the reaction.

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