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3 years ago

\ce{Zn + CuCl2 -> ZnCl2 + Cu}Zn+CuCl2 ZnCl2 +Cu

Chemistry

1 answer:

Kruka [31]3 years ago

3 0

Answer:

0.321 mole of ZnCl₂ have been produced from 21 g of Zn

Explanation:

The reaction is this one:

Zn (s) + CuCl₂ (aq) → ZnCl₂ (aq) + Cu(s)

A redox one.

If the CuCl₂ is available in excess, we consider the limiting reactant, Zn.

Molar mass Zn = 65.38 g/m

Mass / Molar mass = Mol

21 g/ 65.38 g/m = 0.321 mole

As ratio is 1:1, I will produce the same amount of mole, so I'll make 0.321 mole of ZnCl₂

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3 years ago

Given the balanced equation: 2KClO₃ ---> 2KCl + 3O₂ How many moles of O₂ are produced when 4.0 moles of KCl are produced?

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Answer:

The answer to your question is 6.0 moles of O₂

Explanation:

Data

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Process

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3 years ago

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3 years ago

\ce{Zn + CuCl2 -> ZnCl2 + Cu}Zn+CuCl2 ​ ​ ZnCl2 ​ +Cu

\ce{Zn + CuCl2 -> ZnCl2 + Cu}Zn+CuCl2 ZnCl2 +Cu