JEE MAIN - Chemistry (2021 - 24th February Evening Shift - No. 15)
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
8
Explanation
Combustion reaction :
$${C_x}{H_y}(g) + \left( {x + {y \over 4}} \right){O_2}(g) \to xC{O_2}(g) + {y \over 2}{H_2}O(l)$$
Suppose, volume of CxHy is V and volume of O2 is 6 times greater than CxHy = 6V
then volume of xCO2 $$\Rightarrow$$ Vx = 4 V
x = 4
Since, $${V_{{O_2}}} = 6 \times {V_{{C_x}{H_y}}}$$
$$V\left( {x + {y \over 4}} \right)$$ = 6V
$$\left( {x + {y \over 4}} \right) = 6$$ ..... (i)
Put value of x = 4 in Eq. (i) we get,
$$4 + {y \over 4} = 6 \Rightarrow y = 8$$
$${C_x}{H_y}(g) + \left( {x + {y \over 4}} \right){O_2}(g) \to xC{O_2}(g) + {y \over 2}{H_2}O(l)$$
Suppose, volume of CxHy is V and volume of O2 is 6 times greater than CxHy = 6V
then volume of xCO2 $$\Rightarrow$$ Vx = 4 V
x = 4
Since, $${V_{{O_2}}} = 6 \times {V_{{C_x}{H_y}}}$$
$$V\left( {x + {y \over 4}} \right)$$ = 6V
$$\left( {x + {y \over 4}} \right) = 6$$ ..... (i)
Put value of x = 4 in Eq. (i) we get,
$$4 + {y \over 4} = 6 \Rightarrow y = 8$$
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