Combustion of hydrocarbons such as methane (CH_4) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's 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. Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous methane into gaseous carbon dioxide and gaseous water CH_4 (g) + 2O_2 (g) rightarrow CO_2 (g) + 2H_2O (g) Suppose 0.300 kg of methane are burned in air at a pressure of exactly 1 atm and a temperature of 15.0 degree C Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

Question

Chemistry

Sanford

9 January, 22:04

Combustion of hydrocarbons such as methane (CH_4) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's 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. Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous methane into gaseous carbon dioxide and gaseous water CH_4 (g) + 2O_2 (g) rightarrow CO_2 (g) + 2H_2O (g) Suppose 0.300 kg of methane are burned in air at a pressure of exactly 1 atm and a temperature of 15.0 degree C Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

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Huff · Answer 1

443 L of carbon dioxide

Explanation:

Combustion reaction: CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

We assume, that the oxygen is the excess reagent, so let's convert the mass of methane to moles. (mass (g) / molar mass)

Firstly we need to convert the mass from kg to g → 0.300 kg. 1000 g / 1kg

300 g / 16 g/mol = 18.75 moles of methane.

Ratio is 1:1, so 18.75 moles of methane will produce 18.75 moles of CO₂

We apply the Ideal Gases Law to find the answer:

Firstly we need to convert the T°C to T°K → 15°C + 273 = 288 K

P. V = n. R. T → V = (n. R. T) / P We replace dа ta:

V = (18.75 mol. 0.0821 L. atm/mol. K. 288K) / 1 atm → 442.8 ≅ 443 L