JEE MAIN - Chemistry (2021 - 31st August Evening Shift - No. 21)
The transformation occurring in Duma's method is given below :
$${C_2}{H_7}N + \left( {2x + {y \over 2}} \right)CuO \to xC{O_2} + {y \over 2}{H_2}O + {z \over 2}{N_2} + \left( {2x + {y \over 2}} \right)Cu$$
The value of y is ______________. (Integer answer)
$${C_2}{H_7}N + \left( {2x + {y \over 2}} \right)CuO \to xC{O_2} + {y \over 2}{H_2}O + {z \over 2}{N_2} + \left( {2x + {y \over 2}} \right)Cu$$
The value of y is ______________. (Integer answer)
Answer
7
Explanation
$${C_2}{H_7}N + \left( {2x + {y \over 2}} \right)CuO \to xC{O_2} + {y \over 2}{H_2}O + {z \over 2}{N_2} + \left( {2x + {y \over 2}} \right)Cu$$
On balancing
$${C_2}{H_7}N + {{15} \over 2}CuO \to 2C{O_2} + {7 \over 2}{H_2}O + {1 \over 2}{N_2} + {{15} \over 2}Cu$$
On comparing
y = 7
On balancing
$${C_2}{H_7}N + {{15} \over 2}CuO \to 2C{O_2} + {7 \over 2}{H_2}O + {1 \over 2}{N_2} + {{15} \over 2}Cu$$
On comparing
y = 7
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