According to equation 1 mole Zn requires 1 mole of HCl.
First calculate moles of 2.5 g Zn,
As 65.38 g of Zn = 1 mole
Then,
2.5 g equals = X moles
Solving for X,
X = (2.5 g × 1 mol) ÷ 65.38 g
X = 0.038 moles of Zn
Now Calculating moles of HCl,
As 34.46 g of HCl = 1 mole
Then,
5.0 g equals = X moles
Solving for X,
X = (5.0 g × 1 mol) ÷ 34.46 g
X = 0.145 moles of HCl
So,
HCl is provided in excess and Zn is the limiting reactant.
Now Calculating for amount of ZnCl₂ produced. As Zn is limiting reactant so it will control the amount of product produced.
As,
1 mole of Zn produced = 136.28 g of ZnCl₂
Then,
0.038 moles of Zn will produce = X g of ZnCl₂
Solving for X,
X = (0.038 mol × 136.28 g) ÷ 1 mol
X = 5.17 g of ZnCl₂
