To determine the volume of carbon dioxide (CO₂) gas released at standard temperature and pressure (STP) when 6.00 grams of carbon is burned completely, follow these steps:
1. Identify the balanced chemical equation:
The combustion of carbon can be represented by the equation:
[tex]\[
\text{C} + \text{O}_2 \rightarrow \text{CO}_2
\][/tex]
This indicates that 1 mole of carbon (C) combines with 1 mole of oxygen (O₂) to produce 1 mole of carbon dioxide (CO₂).
2. Calculate the molar mass of carbon (C):
The molar mass of carbon (C) is:
[tex]\[
12.01 \text{ grams per mole}
\][/tex]
3. Determine the number of moles of carbon in 6.00 grams:
Use the formula:
[tex]\[
\text{moles of carbon} = \frac{\text{mass of carbon}}{\text{molar mass of carbon}}
\][/tex]
Plugging in the values:
[tex]\[
\text{moles of carbon} = \frac{6.00 \text{ grams}}{12.01 \text{ grams per mole}} \approx 0.4996 \text{ moles}
\][/tex]
4. Understand the stoichiometry of the reaction:
According to the balanced equation, 1 mole of carbon produces 1 mole of CO₂. Therefore, 0.4996 moles of carbon will produce the same number of moles of CO₂.
5. Calculate the volume of CO₂ gas produced at STP:
At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters.
[tex]\[
\text{volume of CO}_2 = \text{moles of CO}_2 \times 22.4 \text{ liters per mole}
\][/tex]
Substituting the values:
[tex]\[
\text{volume of CO}_2 = 0.4996 \times 22.4 \approx 11.19 \text{ liters}
\][/tex]
6. Determine the closest volume option:
By comparing the calculated volume (11.19 liters) with the given options:
[tex]\[
6.0 \text{ L}, \quad 11.2 \text{ L}, \quad 22.4 \text{ L}, \quad 134.4 \text{ L}
\][/tex]
The closest value to 11.19 liters is 11.2 liters.
Therefore, when 6.00 grams of carbon is burned completely, the volume of carbon dioxide gas released at STP is 11.2 liters.
