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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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The pOH of a solution is 9.70. what is the H+ concentration in the solution ?

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Answer:

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Explanation:

We know that for any solution,

pOH + pH = 14

Given,

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pH = 14 - pOH,

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Therefore,

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Here, 2 moles of O-F bond and 2 moles of of O-H bond are broken

1 mol of O=O and 2 moles of H-F bonds are formed.

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4 years ago

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

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