Question

A gaseous fuel mixture stored at 747 mmHg and 298 K contains only methane (CH4) and propane (C3H8). When 11.1 L of this fuel mixture is burned, it produces 780 kJ of heat. What is the mole fraction of methane in the mixture

Answer 1

Answer:

$M_f=38.8\%$

Explanation:

From the question we are told that:

Pressure $P=747mmHg$

Temperature $T=298K$

Volume $V=11.1$

Heat Produced $Q=780kJ$

Generally the equation for ideal gas is mathematically given by

 $PV=nRT$

 $n= (747/760) *11.1/ (0.0821*298)$

 $n=0.446mol$

Therefore

 $x+y=0.446$

 $x=0.446-y .....1$

Since

Heat of combustion of Methane=889 kJ/mol

Heat of combustion of Propane=2220 kJ/mol

Therefore

 $x(889) + y(2220) = 760 ...... 2$

Comparing Equation 1 and 2 and solving simultaneously

 $x=0.446-y .....1$

 $x(889) + y(2220) = 760 ...... 2$

 $x=0.173$

 $y=0.273$

Therefore

Mole fraction 0f Methane is mathematically given as

 $M_f=\frac{x}{n}*100\%$

 $M_f=\frac{1.173}{0.446}*100\%$

 $M_f=38.8\%$