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A sample of .148g of gaseous hydrocarbon (CnH2n+2) reacts completely with oxygen to form CO2 and H2O.?
A sample of 0.148 g of a gaseous hydrocarbon having the chemical formula CnH2n+2 reacts completely with an excess of oxygen gas and carbon dioxide and water. The reaction takes place in a 0.400 L container. Before the reaction, the pressure and temperature in the container are 1.8323 atm and 0 ºC, respectively. After the reaction, the pressure and temperature are 2.7957 atm and 100 ºC, respectively. Find the chemical formula of the hydrocarbon.
2 Answers
- TonyLv 410 years agoFavorite Answer
Hydrocarbon combustion normally follows the following formula:
zCxHy + z(x + y/4)O2 --> zxCO2 + z(y/2)H2O
PV = nRT ==> n = PV/RT
n1 = (1.8323)(0.400)/(0.082)(273) ==> n1 = 2440 mol
n2 = (2.7957)(0.400)/(0.082)(373) ==> n2 = 5087 mol
n1 = 2440 ==> 2440 = z[(x + y/4) + 1]
n2 = 5087 ==> 5087 = z(x + y/2)
m = 0.148 ==> 0.148 = z(12.01x + 1.008y)
Solve the system of equations. That should hopefully be the answer.
- ?Lv 44 years ago
The chemical formula for methane is CH4. Methane is an organic and organic compound. it is likewise categorized as hydrocarbon because of the fact it in straight forward terms includes carbon and hydrogen atoms. while methane is burned, it reacts with oxygen to sort carbon dioxide and water. the generic equation for the reaction is; CH4(g) + O2(g) ------------ > CO2(g) + H2O(g) The balanced equation for the reaction is; CH4(g) + 2O2(g) ------------ > CO2(g) + 2H2O(g) One mole of methane reacts with 2 moles of oxygen molecules to yield one mole of carbon dioxide and a pair of moles of water. or 18.0.5 gm water