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Stoichiometric Calculations

An organic co...

Question

An organic compound is burnt with excess of $O_{2}$ to produce $C O_{2}$ (g) and $H_{2} O (l)$ , which results in 25% volume contraction. Which of the following option(s) satisfy the given conditions.

10 m l C_{3} H_{8} + 110 m l O_{2}

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20 m l C_{3} H_{6} + 80 m l O_{2}

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10 m l C_{3} H_{6} O_{2} + 50 m l O_{2}

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40 m l C_{2} H_{2} O_{4} + 60 m l O_{2}

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Solution

The correct options are
A $10 m l C_{3} H_{8} + 110 m l O_{2}$
C $10 m l C_{3} H_{6} O_{2} + 50 m l O_{2}$
(A) $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (l)$
10 ml propane will react with 50 ml oxygen to form 30 ml carbon dioxide. The volume contraction is $50 + 10 - 30 = 30 m l$ . It is equal to $\frac{30}{10 + 110} \times 100 = 25$ %. Thus option A is correct.
(B) $C_{3} H_{6} (g) + \frac{9}{2} O_{2} (g) \to 3 C O_{2} (g) + 3 H_{2} O (l)$
20 ml propane will react with 90 ml oxygen to form 60 ml carbon dioxide. The volume contraction is $20 + 90 - 60 = 50 m l$ . It is equal to $\frac{50}{20 + 80} \times 100 = 50$ %. Thus option B is incorrect.
(C) $C_{3} H_{6} O_{2} (g) + \frac{7}{2} O_{2} (g) \to 3 C O_{2} (g) + 3 H_{2} O (l)$
10 ml $C_{3} H_{6} O_{2}$ will react with 35 ml oxygen to form 30 ml carbon dioxide. The volume contraction is $0 + 35 - 30 = 15 m l$ . It is equal to $\frac{15}{10 + 50} \times 100 = 25$ %. Thus option C is correct.
(D) $C_{2} H_{2} O_{4} (g) + \frac{1}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l)$
40 ml $C_{3} H_{6} O_{2}$ will react with 20 ml oxygen to form 80 ml carbon dioxide. There is volume expansion. Thus option D is correct.

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