The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O2 for complete oxidation and produces 4 times its own volume of CO2 is CxHy. The value of y is _____________.
Answer (integer)
8
Solution
Combustion reaction :<br/><br/>$${C_x}{H_y}(g) + \left( {x + {y \over 4}} \right){O_2}(g) \to xC{O_2}(g) + {y \over 2}{H_2}O(l)$$<br/><br/>Suppose, volume of C<sub>x</sub>H<sub>y</sub> is V and volume of O<sub>2</sub> is 6 times greater than C<sub>x</sub>H<sub>y</sub> = 6V<br/><br/>then volume of xCO<sub>2</sub> $\Rightarrow$ V<sub>x</sub> = 4 V<br/><br/>x = 4<br/><br/>Since, ${V_{{O_2}}} = 6 \times {V_{{C_x}{H_y}}}$<br/><br/>$V\left( {x + {y \over 4}} \right)$ = 6V<br/><br/>$\left( {x + {y \over 4}} \right) = 6$ ..... (i)<br/><br/>Put value of x = 4 in Eq. (i) we get,<br/><br/>$4 + {y \over 4} = 6 \Rightarrow y = 8$
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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