Question

Combustion of hydrocarbons such as methane (CH) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earths atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide. 1. Write a balanced chemical equation, including physical state symbols, for the iston of ous methane ito gaseous carbon donde and gaseous water 2. Suppose 0.500 kg of methane are burned in air at a pressure of exactly 1 atm and temperatue of 13.0 °C. Caoate the volume of carbon dode gas that is produced. Be sure your answer has the correct number of significant digits

Answer 1

Answer:

The balanced chemical reaction of combustion of methane is:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

734 liters of carbon dioxide gas is produced. .

Explanation:

The balanced chemical reaction of combustion of methane is:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

Mass of methane gas = 0.500 kg = 500 g (1 kg= 1000 g)

Moles of methane = $\frac{500 g}{16 g/mol}=31.25 mol$

According to reaction, 1 mole of methane gas gives 1 mole of carbon dioxide gas. Then 31.25 moles of methane will give :

$\frac{1}{1}\times 31.25 mol=31.25 mol$ of carbon dioxide

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = $\1 atm$

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:

$V=\frac{31.25\times 0.0821 atm L/mol K\times 286.15 K}{1 atm}$

V = 734.15 L ≈ 734 L

734 liters of carbon dioxide gas is produced. .

Answer 2

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