Answer to Question #150923 in General Chemistry for AA

Question

A mixture of propane and methane occupied 122 mL at 25⁰C and 1.0 atm. This mixture was totally burned in excess O2 to produce CO2(g) and H2O(g). The mass of the CO2 produced was 0.506 g. What is the mass % composition of the original mixture? Show all work.

Answer 1

CH4 + 2O2 = CO2 + 2H2O.

C3H8 + 5O2 = 3CO2 + 4H2O.

pV=nRT. n= pV/RT = 1 x 0.122 / 0.082 x 298 = 0.005.

x+y = 0.005.

0.506/44= 0.0115.

x+3y = 0.0115.

x= 0,00175

y= 0,00325.

0.00175 x 16= 0.028 g of CH4.

0.00325 x 44 = 0.143 g of C3H8.

0.028 x 100/0.028+0.143 = 16,374269 %. of CH4.

0.143x100/0.028+0.143= 83,625731 % of C3H8.