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