\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

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

hope this !

explanation:

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be