Answer:
The balanced chemical reaction of combustion of methane is:
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)[/tex]
734 liters of carbon dioxide gas is produced. .
Explanation:
The balanced chemical reaction of combustion of methane is:
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)[/tex]
Mass of methane gas = 0.500 kg = 500 g (1 kg= 1000 g)
Moles of methane = [tex]\frac{500 g}{16 g/mol}=31.25 mol[/tex]
According to reaction, 1 mole of methane gas gives 1 mole of carbon dioxide gas. Then 31.25 moles of methane will give :
[tex]\frac{1}{1}\times 31.25 mol=31.25 mol[/tex] of carbon dioxide
Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas = [tex]\1 atm[/tex]
V = Volume of gas =?
n = number of moles of carbon dioxide gas = 31.25 mol
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of gas =13.0°C=13.0+273.15 K= 286.15 K
Putting values in above equation, we get:
[tex]V=\frac{31.25\times 0.0821 atm L/mol K\times 286.15 K}{1 atm}[/tex]
V = 734.15 L ≈ 734 L
734 liters of carbon dioxide gas is produced. .
Answer:
Equation if the reaction
CH4(g) + 2O2(g) = CO2(g) + 20k H2O(g)
PV = nRT
P = 1atm, V = ?, T = 13+273 = 682k,
R = 0.0821,
n = 0.500g/19.04g/mol =0.03117mol
One molecule of methane produced one molecule of carbon dioxide
therefore, 0.03117 mole of CH4 will produce 0.03117mol of CO2 the no of mol
V nRT/P
V = 0.03117*0.0821*286/1
V = 0.7319dm3
