If 27g of water is formed during complete combustion of pure propene (`C_(3)H_(8)`), the mass of propene burnt is

Options

A. 42 g
B. 21 g
C. 14 g of `N_(2)`
D. 56 g

Explanation

To find the mass of propene burnt when 27 g of water is formed during its complete combustion, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of propene.
The balanced equation for the combustion of propene (C₃H₈) is:
\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]

### Step 2: Determine the molar masses of the substances involved.
- Molar mass of propene (C₃H₈):
- Carbon (C): 12 g/mol × 3 = 36 g/mol
- Hydrogen (H): 1 g/mol × 8 = 8 g/mol
- Total: 36 + 8 = 44 g/mol

- Molar mass of water (H₂O):
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Oxygen (O): 16 g/mol × 1 = 16 g/mol
- Total: 2 + 16 = 18 g/mol

### Step 3: Use stoichiometry to relate the mass of water to the mass of propene.
From the balanced equation, we see that 1 mole of propene produces 4 moles of water.

- Moles of water produced from 27 g:
\[ \text{Moles of } H_2O = \frac{27 \text{ g}}{18 \text{ g/mol}} = 1.5 \text{ moles} \]

### Step 4: Calculate the moles of propene required to produce 1.5 moles of water.
From the balanced equation, 1 mole of propene produces 4 moles of water. Therefore, to find the moles of propene needed for 1.5 moles of water:
\[ \text{Moles of } C_3H_8 = \frac{1.5 \text{ moles } H_2O}{4} = 0.375 \text{ moles} \]

### Step 5: Convert moles of propene to grams.
Now, we convert the moles of propene to grams using its molar mass:
\[ \text{Mass of } C_3H_8 = 0.375 \text{ moles} \times 44 \text{ g/mol} = 16.5 \text{ g} \]

### Final Answer:
The mass of propene burnt is **16.5 grams**.

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