Answer:
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
