How many carbon dioxide and water molecules will be there in the product side, if the following equation is made balanced? C 2 H 5 OH+O 2 →CO 2 +H 2 O
- (a)3 and 6, respectively
- (b)1 and 1, respectively
- (c)2 and 3, respectively
- (d)3 and 2, respectively
Answer
Why
Correct — C. Balance carbon first, then hydrogen, then oxygen.
C: the left side has 2 carbons in C₂H₅OH → 2 CO₂
H: the left side has 6 hydrogens, 5 plus 1 in the OH → 3 H₂O
O on the right: 2×2 + 3 = 7 oxygens
O on the left: 1 (from OH) + 2n = 7, so n = 3 → 3 O₂
Balanced, this reads C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. The product side carries 2 and 3 — option (c).
Why the others are wrong
- (a)3 and 6, respectively — 3 CO₂ demands three carbon atoms and ethanol supplies only two. 6 H₂O would need twelve hydrogens against the six available.
- (b)1 and 1, respectively — 1 and 1 leaves one carbon and four hydrogens stranded on the left with nowhere to go. It balances neither element.
- (d)3 and 2, respectively — 3 and 2 inverts the correct pair and again asks ethanol for a third carbon. The water coefficient is fixed at 3 by the six hydrogens.
Concept
A balanced equation is an accounting statement: every atom on the left reappears on the right, because mass is conserved in a chemical change.
The order of work matters. Balance the element that appears in one compound on each side first — here carbon, then hydrogen — and leave oxygen for last, because free O₂ can absorb whatever coefficient the rest of the equation demands.
Complete combustion of an alcohol or a hydrocarbon yields only CO₂ and H₂O, so the two product coefficients are settled by the carbon and hydrogen counts alone.
The stem prints the formula as C 2 H 5 OH with spaces, which is how the response sheet renders subscripts. Read it as C₂H₅OH.
Key facts
- The balanced combustion of ethanol is C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O.
- Ethanol, C₂H₅OH, carries 2 carbon atoms and 6 hydrogen atoms, one of the six sitting in the OH group.
- Balancing enforces the law of conservation of mass: atoms are neither created nor destroyed in a reaction.
Study next
Common traps
- Forgetting the hydrogen inside the OH group and counting only five.
- Balancing oxygen before hydrogen, then having to redo the whole equation.
- Answering with the O₂ coefficient of 3 when the question asks only about the product side.
SSC slips school chemistry into General Awareness as a one-step calculation rather than pure recall. The equation is printed unbalanced and only the two product coefficients are wanted, so the oxygen coefficient never has to be reported.
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