How many gm of bromine will react with 21 gm `C_3H_6`

Options

A. 80
B. 160
C. 240
D. 320

Explanation

To determine how many grams of bromine will react with 21 grams of propene (C₃H₆), we can follow these steps:

### Step 1: Identify the Reaction
Propene (C₃H₆) reacts with bromine (Br₂) in an addition reaction. The balanced chemical equation for the reaction is:
\[ \text{C}_3\text{H}_6 + \text{Br}_2 \rightarrow \text{C}_3\text{H}_6\text{Br}_2 \]

### Step 2: Calculate the Molar Mass of C₃H₆
The molar mass of propene (C₃H₆) can be calculated as follows:
- Carbon (C) has a molar mass of approximately 12 g/mol.
- Hydrogen (H) has a molar mass of approximately 1 g/mol.

Calculating the molar mass:
\[ \text{Molar mass of C}_3\text{H}_6 = (3 \times 12) + (6 \times 1) = 36 + 6 = 42 \text{ g/mol} \]

### Step 3: Calculate the Molar Mass of Br₂
The molar mass of bromine (Br₂) is:
- Bromine (Br) has a molar mass of approximately 80 g/mol.

Calculating the molar mass:
\[ \text{Molar mass of Br}_2 = 2 \times 80 = 160 \text{ g/mol} \]

### Step 4: Set Up the Stoichiometric Ratio
From the balanced equation, 1 mole of C₃H₆ reacts with 1 mole of Br₂. Therefore, the ratio of C₃H₆ to Br₂ is 1:1.

### Step 5: Calculate the Amount of Br₂ Needed for 21 g of C₃H₆
Using the stoichiometric ratio, we can find out how much bromine is needed for 21 grams of propene.

First, calculate the number of moles of C₃H₆ in 21 grams:
\[ \text{Number of moles of C}_3\text{H}_6 = \frac{\text{mass}}{\text{molar mass}} = \frac{21 \text{ g}}{42 \text{ g/mol}} = 0.5 \text{ moles} \]

Since the ratio of C₃H₆ to Br₂ is 1:1, 0.5 moles of C₃H₆ will react with 0.5 moles of Br₂.

Now, calculate the mass of Br₂ needed:
\[ \text{Mass of Br}_2 = \text{Number of moles} \times \text{Molar mass} = 0.5 \text{ moles} \times 160 \text{ g/mol} = 80 \text{ g} \]

### Final Answer
Thus, 80 grams of bromine will react with 21 grams of C₃H₆.

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