A gaseous hydrocarbon requires 6 times its own volume of -O-2- complete oxidation and produces 4 times its volume of C-O-2- same pressure and temperature- The formula of gas is -C-4-H-10-C-4-H-8-C-4-H-6-C-8-H-16-

Question

Toppr Admin · Accepted Answer

The balanced reaction is as follows-xA0-CXHY-X-Y4-O2-x2192-XCO2-Y2H2OLet- a litres of CXHY be used-xA0-a-X-Y4-6aand given- aX-4aTherefore- X-4 and Y-8Hence- the gas is C4H8-

A gaseous hydrocarbon requires $6$ times its own volume of $O_{2}$ for complete oxidation and produces $4$ times its volume of $C O_{2}$ at same pressure and temperature. The formula of gas is :
$C_{4} H_{10}$
$C_{4} H_{8}$
$C_{4} H_{6}$
$C_{8} H_{16}$

The balanced reaction is as follows:
$C_{X} H_{Y} + [X + \frac{Y}{4}] O_{2} \to X C O_{2} + \frac{Y}{2} H_{2} O$
Let, $a$ litres of $C_{X} H_{Y}$ be used.
$a [X + \frac{Y}{4}] = 6 a$
and given, $a X = 4 a$
Therefore, $X = 4$ and $Y = 8$
Hence, the gas is $C_{4} H_{8}$ .

A gaseous hydrocarbon requires 6 times its own volume of O2 for complete oxidation and produces 4 times its volume of CO2 at same pressure and temperature. The formula of gas is :C4H10C4H8C4H6C8H16

The balanced reaction is as follows: CXHY+[X+Y4]O2→XCO2+Y2H2OLet, a litres of CXHY be used. a[X+Y4]=6aand given, aX=4aTherefore, X=4 and Y=8Hence, the gas is C4H8.

A gaseous hydrocarbon requires $6$ times its own volume of $O_{2}$ for complete oxidation and produces $4$ times its volume of $C O_{2}$ at same pressure and temperature. The formula of gas is :
$C_{4} H_{10}$
$C_{4} H_{8}$
$C_{4} H_{6}$
$C_{8} H_{16}$

The balanced reaction is as follows:
$C_{X} H_{Y} + [X + \frac{Y}{4}] O_{2} \to X C O_{2} + \frac{Y}{2} H_{2} O$
Let, $a$ litres of $C_{X} H_{Y}$ be used.
$a [X + \frac{Y}{4}] = 6 a$
and given, $a X = 4 a$
Therefore, $X = 4$ and $Y = 8$
Hence, the gas is $C_{4} H_{8}$ .