How much gm of bromine reacts with 3.15 gm of butane -2-ene (Atomic mass of bromine`=80`)

Explanation

To determine how much bromine reacts with 3.15 grams of butane-2-ene, we will follow these steps:

### Step 1: Determine the molar mass of butane-2-ene (C4H8)
The molecular formula of butane-2-ene is C4H8. To find its molar mass, we add the atomic masses of all the atoms in the formula:
- Carbon (C) has an atomic mass of approximately 12 g/mol.
- Hydrogen (H) has an atomic mass of approximately 1 g/mol.

Calculating the molar mass:
\[ \text{Molar mass of butane-2-ene} = (4 \times 12) + (8 \times 1) = 48 + 8 = 56 \text{ g/mol} \]

### Step 2: Determine the molar mass of bromine (Br2)
Bromine (Br) has an atomic mass of 80 g/mol. Since bromine exists as a diatomic molecule (Br2), we calculate its molar mass as follows:
\[ \text{Molar mass of Br2} = 2 \times 80 = 160 \text{ g/mol} \]

### Step 3: Establish the stoichiometric relationship
From the balanced chemical reaction, we know that:
\[ 1 \text{ mol of butane-2-ene reacts with 1 \text{ mol of Br2}} \]
This means that 56 g of butane-2-ene reacts with 160 g of bromine.

### Step 4: Calculate the amount of bromine that reacts with 3.15 g of butane-2-ene
Using the ratio from the molar masses:
\[ \text{If } 56 \text{ g of butane-2-ene reacts with } 160 \text{ g of bromine, then } 3.15 \text{ g of butane-2-ene will react with } x \text{ g of bromine.} \]
We can set up a proportion:
\[ \frac{56 \text{ g butane-2-ene}}{160 \text{ g bromine}} = \frac{3.15 \text{ g butane-2-ene}}{x \text{ g bromine}} \]

Cross-multiplying gives:
\[ 56x = 3.15 \times 160 \]

Calculating the right side:
\[ 3.15 \times 160 = 504 \]
Now, solving for \(x\):
\[ x = \frac{504}{56} = 9 \text{ g} \]

### Final Answer
The amount of bromine that reacts with 3.15 grams of butane-2-ene is **9 grams**.

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