Question

A sealed chamber contains 9.00 g CH₄ and 15.00 g O₂. The mixture is ignited. How many grams of CO₂ are produced?

Answer 1

The balanced chemical equation for the combustion of methane (CH₄) in oxygen (O₂) yields one molecule of carbon dioxide (CO₂) and two molecules of water (H₂O). Given the quantities of methane and oxygen present in the chamber, the reaction produces 10.30 grams of carbon dioxide.

Step-by-step explanation:

The balanced chemical equation for the combustion of methane is:

`\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]`

To find the grams of CO₂ produced, first, determine the limiting reactant. Calculate the moles of CH₄ and O₂:

`Moles of CH₄ = \(\frac{9.00 \text{ g}}{16.04 \text{ g/mol}} = 0.561 \text{ mol}\)`

`Moles of O₂ = \(\frac{15.00 \text{ g}}{32.00 \text{ g/mol}} = 0.469 \text{ mol}\)`

According to the balanced equation, the ratio of moles of CH₄ to moles of O₂ is 1:2. Thus, O₂ is the limiting reactant since it will be completely consumed before CH₄.

Next, determine the moles of CO₂ produced based on the moles of O₂ used:

`Moles of CO₂ = \(0.469 \text{ mol} * \frac{1 \text{ mol CO}_2}{2 \text{ mol O}_2} = 0.234 \text{ mol CO}_2\)`

Finally, find the grams of CO₂ produced:

`Grams of CO₂ = \(0.234 \text{ mol} * 44.01 \text{ g/mol} = 10.30 \text{ g CO}_2\)`

Therefore, the reaction produces 10.30 grams of CO₂.

This calculation assumes complete combustion and ideal conditions. Real reactions may yield varying results due to factors like incomplete combustion or impurities in the reactants.