Question

combustion of hydrocarbons such as methane (ch₄) produces carbon dioxide, a \greenhouse gas.\ greenhouse gases in the earths atmosphere can trap the suns 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 combustion of gaseous methane into gaseous carbon dioxide and gaseous water. 2. suppose 0.350 kg of methane are burned in air at a pressure of exactly 1 atm and a temperature of 12.0 °c. calculate the volume of carbon dioxide gas that is produced. be sure your answer has the correct number of significant digits. l

Explanation:

1.

Step1: Write the un - balanced equation

Methane (\(CH_{4}(g)\)) reacts with oxygen (\(O_{2}(g)\)) to form carbon dioxide (\(CO_{2}(g)\)) and water (\(H_{2}O(g)\)). The un - balanced equation is \(CH_{4}(g)+O_{2}(g)\to CO_{2}(g)+H_{2}O(g)\).

Step2: Balance the hydrogen atoms

There are 4 hydrogen atoms in \(CH_{4}\). To balance the hydrogen atoms, we put a 2 in front of \(H_{2}O\). The equation becomes \(CH_{4}(g)+O_{2}(g)\to CO_{2}(g)+2H_{2}O(g)\).

Step3: Balance the oxygen atoms

Now, for the oxygen atoms: on the right - hand side, there are \(2 + 2=4\) oxygen atoms (2 from \(CO_{2}\) and 2 from \(2H_{2}O\)). So we put a 2 in front of \(O_{2}\).

Step1: Calculate the number of moles of methane

The molar mass of \(CH_{4}\) is \(M=(12 + 4\times1)\space g/mol=16\space g/mol\). The mass of methane \(m = 0.350\space kg=350\space g\). Using the formula \(n=\frac{m}{M}\), we have \(n_{CH_{4}}=\frac{350\space g}{16\space g/mol}=\frac{350}{16}\space mol\approx21.875\space mol\).

Step2: Use the stoichiometry of the balanced equation

From the balanced equation \(CH_{4}(g)+2O_{2}(g)=CO_{2}(g)+2H_{2}O(g)\), the mole ratio of \(CH_{4}\) to \(CO_{2}\) is \(1:1\). So \(n_{CO_{2}}=n_{CH_{4}}\approx21.875\space mol\).

Step3: Use the ideal gas law \(PV = nRT\)

We are given \(P = 1\space atm\), \(T=(12.0 + 273.15)\space K=285.15\space K\), and \(R = 0.0821\space L\cdot atm/(mol\cdot K)\). Rearranging the ideal gas law for \(V\), we get \(V=\frac{nRT}{P}\).
Substitute \(n = 21.875\space mol\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 285.15\space K\), and \(P = 1\space atm\) into the formula:
\(V=\frac{21.875\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times285.15\space K}{1\space atm}\)
\(V=21.875\times0.0821\times285.15\space L\)
\(V\approx21.875\times0.0821\times285\space L\) (approximate \(285.15\approx285\) for simplicity in calculation)
\(V\approx21.875\times23.4\space L\) (since \(0.0821\times285\approx23.4\))
\(V\approx512\space L\)

Answer:

\(CH_{4}(g)+2O_{2}(g)=CO_{2}(g)+2H_{2}O(g)\)

2.