Answer:
The balanced chemical reaction of combustion of methane is:
![CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)](https://img.qammunity.org/2020/formulas/chemistry/college/c95smvmeqfz6ofnqhcvlf3pfojv1ffbk3x.png)
734 liters of carbon dioxide gas is produced. .
Step-by-step explanation:
The balanced chemical reaction of combustion of methane is:
![CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)](https://img.qammunity.org/2020/formulas/chemistry/college/c95smvmeqfz6ofnqhcvlf3pfojv1ffbk3x.png)
Mass of methane gas = 0.500 kg = 500 g (1 kg= 1000 g)
Moles of methane =
![(500 g)/(16 g/mol)=31.25 mol](https://img.qammunity.org/2020/formulas/chemistry/college/j6ijvv35yoecik05ebbqe2v9z30w8n72c9.png)
According to reaction, 1 mole of methane gas gives 1 mole of carbon dioxide gas. Then 31.25 moles of methane will give :
of carbon dioxide
Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas =
![\1 atm](https://img.qammunity.org/2020/formulas/chemistry/college/hjniq9mw8evu0beonlbne08sp2er0bnptw.png)
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:
![V=(31.25* 0.0821 atm L/mol K* 286.15 K)/(1 atm)](https://img.qammunity.org/2020/formulas/chemistry/college/2c1rugugrccdaf3sgron9yrz08gd74smyf.png)
V = 734.15 L ≈ 734 L
734 liters of carbon dioxide gas is produced. .