2.50 g of zn is added to 5.00 g of hcl

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

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Naoh + hbr -> nabr (salt) + h2o ca (oh)2 + 2hbr - - - > ca (br)2 + 2h2o