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\ce{Zn + CuCl2 -> ZnCl2 + Cu}Zn+CuCl2 ​ ​ ZnCl2 ​ +Cu
How many moles of \ce{ZnCl2}ZnClX 2 ​ will be produced from 21.0 \text{ g}21.0 g21, point, 0, start text, space, g, end text of \ce{Zn}ZnZ, n, assuming \ce{CuCl2}CuClX 2 ​ is available in excess

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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

Eduard22sly

The number of mole of ZnCl₂ obtained when 21 g of Zn react with excess CuCl₂ is 0.323 mole

How to determine the mole of Zn

  • Mass of Zn = 21 g
  • Molar mass of Zn = 65 g/mol
  • Mole of Zn =?

Mole = mass / molar mass

Mole of Zn = 21 / 65

Mole of Zn = 0.323 mole

How to determine the mole of ZnCl₂ produced

Balanced equation

Zn + CuCl₂ —> ZnCl₂ + Cu

From the balanced equation above,

1 mole of Zn reacted to produce 1 mole of ZnCl₂.

Therefore,

0.323 mole of Zn will also react to produce 0.323 mole of ZnCl₂

Thus, 0.323 mole of ZnCl₂ was obtained from the reaction

Learn more about stoichiometry:

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