\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

a is the correct answer