Answer:
A. HCl is the limiting reactant.
B. Theoretical yield of Cl₂ is 20.9 g.
C. Actual yield of Cl₂ = 15.2 g.
Step-by-step explanation:
The balanced equation for the reaction is given below:
4HCl + MnO₂ –> MnCl₂ + 2H₂O + Cl₂
Next, we shall determine the masses of HCl and MnO₂ that reacted and the mass of Cl₂ produced from the balanced equation. This can be obtained as follow:
Molar mass of HCl = 1 + 35.5
= 36.5 g/mol
Mass of HCl from the balanced equation = 4 × 36.5 = 146 g
Molar mass of MnO₂ = 55 + (2×16)
= 55 + 32
= 87 g/mol
Mass of MnO₂ from the balanced equation = 1 × 87 = 87 g
Molar mass of Cl₂ = 2 × 35.5
= 71 g/mol
Mass of Cl₂ from the balanced equation = 1 × 71 = 71 g
SUMMARY:
From the balanced equation above,
146 g of HCl reacted with 87 g of MnO₂ to produce 71 g of Cl₂.
A. Determination of the limiting reactant.
From the balanced equation above,
146 g of HCl reacted with 87 g of MnO₂.
Therefore, 42.9 g of HCl will react with = (42.9 × 87)/146 = 25.6 g of MnO₂.
From the calculation made above, we can see clearly that only 25.6 g out of 43.1 g of MnO₂ given was needed to react completely with 42.9 g of HCl.
Therefore, HCl is the limiting reactant.
B. Determination of theoretical yield of Cl₂.
Here, the limiting reactant will be used.
From the balanced equation above,
146 g of HCl reacted to produce 71 g of Cl₂.
Therefore, 42.9 g of HCl will react to produce = (42.9 × 71)/146 = 20.9 g of Cl₂.
Thus, the theoretical yield of Cl₂ is 20.9 g.
C. Determination of the actual yield of Cl₂.
Theoretical yield of Cl₂ = 20.9 g
Percentage yield of Cl₂ = 72.9%
Actual yield of Cl₂ =?
Percentage yield = Actual yield / Theoretical yield × 100
72.9% = Actual yield / 20.9
Cross multiply
Actual yield = 72.9% × 20.9
Actual yield = 72.9/100 × 20.9
Actual yield = 0.729 × 20.9
Actual yield of Cl₂ = 15.2 g