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aleksley [76]
3 years ago
14

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 mix

ture is burned, it produces 780 kJ of heat. What is the mole fraction of methane in the mixture
Chemistry
1 answer:
Alisiya [41]3 years ago
7 0

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\%

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