Question

In bromination of Propyne, with Bromine 1,1,2,2-tetrabromopropane is obtained in 27% yield. The amount of 1,1,2,2-tetrabromopropane obtained from 1 g of Bromine in this reaction is _______ × 10^–1 g. (Nearest integer)
(Molar Mass : Bromine = 80 g/mol)

Answer 1

Correct Answer: 3

To solve this problem, we will follow these steps: 

1. Calculate the moles of Bromine used.
2. Determine the moles of 1,1,2,2-tetrabromopropane produced based on the yield.
3. Compute the mass of 1,1,2,2-tetrabromopropane formed.
4. Verify the result falls within the specified range.

Step 1: Calculate the moles of Bromine
The molar mass of Br₂ is 160 g/mol (since each Br atom has a mass of 80 g/mol).
Moles of Br₂ = 1 g / 160 g/mol = 0.00625 mol

Step 2: Determine the moles of 1,1,2,2-tetrabromopropane
From the stoichiometry of the reaction, 1 mole of propyne reacts with 2 moles of Bromine to give 1 mole of 1,1,2,2-tetrabromopropane. Thus, the mole ratio of Br₂ to 1,1,2,2-tetrabromopropane is 2:1.
Moles of 1,1,2,2-tetrabromopropane = 0.00625 mol / 2 = 0.003125 mol

Step 3: Compute the mass of 1,1,2,2-tetrabromopropane formed
Molar mass of 1,1,2,2-tetrabromopropane (C₃H₂Br₄):
Carbon (C): 3 × 12 g/mol = 36 g/mol
Hydrogen (H): 2 × 1 g/mol = 2 g/mol
Bromine (Br): 4 × 80 g/mol = 320 g/mol
Total molar mass = 36 + 2 + 320 = 358 g/mol
Mass of 1,1,2,2-tetrabromopropane = 0.003125 mol × 358 g/mol = 1.11875 g
Considering a 27% yield:
Mass obtained = 1.11875 g × 0.27 = 0.3020625 g

Step 4: Verify the result
The question asks for the amount in × 10^–1g. Thus, convert 0.3020625 g accordingly:
0.3020625 g = 3.020625 × 10^–1g
Nearest integer = 3

The computed value is within the specified range of 3. Thus, the amount of 1,1,2,2-tetrabromopropane obtained is 3 × 10^–1 g.

Answer 2

2 moles Br₂=1 mole 1,1,2,2-tetrabromopropane
\(\frac{1}{160} \, \text{mole Br}_2 = \frac{1}{2} \times \frac{1}{160} \, \text{mole 1,1,2,2-tetrabromopropane}\)
But yield of reaction is only 27%
Moles of 1,1,2,2-tetrabromopropane \(=\)\(\frac{1}{2} \times \frac{1}{160} \times \frac{27}{100}\)
Molar mass of 1,1,2,2-tetrabromopropane = 360 g
Mass of 1,1,2,2-tetrabromopropane\(=\)\(12 \times \frac{1}{160} \times \frac{27}{100} \times 360 \, \text{g}\)
\(= 3 × 10^{–1} \text{g}\)
So, the answer is 3.