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 0.490 kg of methane are burned in air at a pressure of exactly 1 atm and a temperature of 10.0 °c. calculate the volume of carbon dioxide gas that is produced. be sure your answer has the correct number of significant digits.

Explanation:

1. Balanced Chemical Equation

Step1: Write the unbalanced 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 unbalanced 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 \(CH_{4}\). So, we put a coefficient of 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, there are 4 oxygen atoms on the right - hand side (2 in \(CO_{2}\) and 2 in \(2H_{2}O\)). So, we put a coefficient of 2 in front of \(O_{2}\).
The balanced equation is \(CH_{4}(g)+2O_{2}(g)=CO_{2}(g)+2H_{2}O(g)\).

Step1: Calculate the number of moles of \(CH_{4}\)

The molar mass of \(CH_{4}\) (\(M\)) is \(M=(12 + 4\times1)\space g/mol=16\space g/mol\). The mass of \(CH_{4}\), \(m = 0.490\space kg=490\space g\).
Using the formula \(n=\frac{m}{M}\), we have \(n_{CH_{4}}=\frac{490\space g}{16\space g/mol}=30.625\space mol\).

Step2: Use the mole ratio from the balanced equation

From the balanced equation \(CH_{4}(g)+2O_{2}(g)=CO_{2}(g)+2H_{2}O(g)\), the mole ratio \(n_{CH_{4}}:n_{CO_{2}} = 1:1\). So, \(n_{CO_{2}}=n_{CH_{4}} = 30.625\space mol\).

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

We are given \(P = 1\space atm\), \(T=(10.0 + 273.15)\space K=283.15\space K\), \(n = 30.625\space mol\), 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 the values: \(V=\frac{30.625\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times283.15\space K}{1\space atm}\).
\(V=\frac{30.625\times0.0821\times283.15}{1}\space L\)
\(V = 30.625\times0.0821\times283.15\space L\)
\(V=30.625\times23.246615\space L\)
\(V = 712\space L\) (rounded to three significant digits)

Answer:

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

2. Volume of \(CO_{2}\) produced