If 35.8 g mg react with 82.3 g hcl according to the reaction below, how many grams of hydrogen gas will produced, and how many grams of the excess reactant will be left over? unbalanced equation: mg + hci —> mgcl2 + h2

             Amount of H₂ Produced =  2.27 g of H₂

             Excess amount of Mg  =  4.6 g of Mg

Solution:

The Balance Chemical Equation is as follow,

                                    Mg  +  2 HCl    →     MgCl₂  +  H₂

Step 1: Find out the Limiting Reagent:

According to equation,

                 24.3 g (1 mol) Mg reacts with  =  72.92 g (2 mol) HCl

So,

                       35.8 g Mg will react with  =  X g of HCl

Solving for X,

                      X  =  (35.8 g × 72.92 g) ÷ 24.3 g

                      X  =  107.42 g of HCl

It means for complete consumption of 35.8 g of Mg we will require 107.42 g of HCl but, we are only provided with 82.3 g of HCl. Hence, HCl is the Limiting reagent and will control the yield of product.

Step 2: Calculating amount of H₂ produced:

According to equation,

                 72.92 g (2 mol) HCl produce  =  2.016 g (1 mol) H₂

So,

                       82.3 g HCl will produce  =  X g of H₂

Solving for X,

                      X  =  (82.3 g × 2.016 g) ÷ 72.92 g

                      X  =  2.27 g of H₂

Step 3: Calculating amount of excess reagent left:

As we know Mg is the access reagent. The amount of it left unreacted is calculated as,

According to equation,

                 72.92 g (2 mol) HCl reacted with  =  35.8 g (1 mol) Mg

So,

                       82.3 g HCl will react with  =  X g of Mg

Solving for X,

                      X  =  (82.3 g × 35.8 g) ÷ 82.3 g

                      X  =  40.40 g of Mg

Therefore,

                   Excess amount of Mg  =  40.40 g - 35.8 g

                   Excess amount of Mg  =  4.6 g of Mg

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