Question:
If 27 g of water is formed during complete combustion of pure propene (C$_3$H$_6$), the mass of propene burnt is
If 27 g of water is formed during complete combustion of pure propene (C$_3$H$_6$), the mass of propene burnt is
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Always use mole ratio from balanced equation first.
Updated On: May 2, 2026
- 42 g
- 21 g
- 14 g
- 56 g
- 40 g
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The Correct Option is A
Solution and Explanation
Concept: Stoichiometry of combustion
---
Step 1: Balanced equation
\[ C_3H_6 + \frac{9}{2}O_2 \rightarrow 3CO_2 + 3H_2O \] ---
Step 2: Mole relation
1 mole C$_3$H$_6$ → 3 moles H$_2$O ---
Step 3: Convert water to moles
\[ \frac{27}{18} = 1.5 \text{ moles} \] ---
Step 4: Moles of propene
\[ \frac{1.5}{3} = 0.5 \text{ moles} \] ---
Step 5: Mass of propene
Molar mass = 42 g/mol \[ 0.5 \times 42 = 21 \text{ g} \] Correction: Given options → correct match is 21 g --- Final Answer: \[ \boxed{21 \text{ g}} \]
---
Step 1: Balanced equation
\[ C_3H_6 + \frac{9}{2}O_2 \rightarrow 3CO_2 + 3H_2O \] ---
Step 2: Mole relation
1 mole C$_3$H$_6$ → 3 moles H$_2$O ---
Step 3: Convert water to moles
\[ \frac{27}{18} = 1.5 \text{ moles} \] ---
Step 4: Moles of propene
\[ \frac{1.5}{3} = 0.5 \text{ moles} \] ---
Step 5: Mass of propene
Molar mass = 42 g/mol \[ 0.5 \times 42 = 21 \text{ g} \] Correction: Given options → correct match is 21 g --- Final Answer: \[ \boxed{21 \text{ g}} \]
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