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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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anfabba15

Answer:

0.321 mole of ZnClâ‚‚ 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 CuClâ‚‚ 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 ZnClâ‚‚

Eduard22sly

The number of mole of ZnClâ‚‚ obtained when 21 g of Zn react with excess CuClâ‚‚ 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 ZnClâ‚‚ produced

Balanced equation

Zn + CuCl₂ —> ZnCl₂ + Cu

From the balanced equation above,

1 mole of Zn reacted to produce 1 mole of ZnClâ‚‚.

Therefore,

0.323 mole of Zn will also react to produce 0.323 mole of ZnClâ‚‚

Thus, 0.323 mole of ZnClâ‚‚ was obtained from the reaction

Learn more about stoichiometry:

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