Answer to Question #149236 in General Chemistry for teacher

Question

If 45.0 L of natural gas, which is essentially methane (CH4), undergoes
complete combustion at 730 mm Hg and 20ºC, how many grams of each
product are formed?

Answer 1

CH₄ + 2 O₂ = CO₂ + 2 H₂O

(45L CH₄) (1LCO₂/ 1L CH₄) = 45L CO₂

PV = n RT. n = (mass/ molar mass)

PV = MRT. M = (PVM/ RT)

M = ((730mmHg)(45)(44g/mol))/(62.4mmHg)(293k) =79.1 =80g CO₂

(45L CH₄ )((2LH₂O/1LCH₄ ) = 90L H₂O

M = PVM/ RT = (730)(90)(18.02)/(42.4L mmHg) (293)

64.8= 65 = 70 H₂O

60g H₂O