To determine the number of milliliters of carbon monoxide (CO) gas that can be produced from 8.00 grams of silica (SiO₂) and 1.00 liter of chlorine gas (Cl₂), we need to follow a detailed step-by-step solution based on the stoichiometry of the reaction. Let's break it down:
1. Balanced Chemical Equation:
[tex]\[
2 \text{Cl}_2 (g) + \text{SiO}_2 (s) + 2 \text{C} (s) \rightarrow \text{SiCl}_4 (l) + 2 \text{CO} (g)
\][/tex]
According to the balanced equation:
- 2 moles of Cl₂ react with 1 mole of SiO₂ to produce 2 moles of CO.
2. Determine the moles of SiO₂ available:
- Molar mass of SiO₂ = 60.09 g/mol
- Given mass of SiO₂ = 8.00 g
[tex]\[
\text{Moles of } \text{SiO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{8.00 \text{ g}}{60.09 \text{ g/mol}} \approx 0.1331 \text{ mol}
\][/tex]
3. Calculate the moles of CO produced:
- From the balanced equation, 1 mole of SiO₂ produces 2 moles of CO.
[tex]\[
\text{Moles of CO} = 2 \times \text{moles of } \text{SiO}_2 = 2 \times 0.1331 \text{ mol} = 0.2663 \text{ mol}
\][/tex]
4. Calculate the volume of CO produced at STP:
- At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters.
[tex]\[
\text{Volume of CO (L)} = \text{moles of CO} \times 22.4 \text{ L/mol} = 0.2663 \text{ mol} \times 22.4 \text{ L/mol} \approx 5.964 \text{ L}
\][/tex]
5. Convert the volume of CO from liters to milliliters:
- There are 1000 milliliters in 1 liter.
[tex]\[
\text{Volume of CO (mL)} = 5.964 \text{ L} \times 1000 \text{ mL/L} = 5964.4 \text{ mL}
\][/tex]
So, the number of milliliters of CO gas that can be produced from 8.00 grams of silica and 1.00 liter of chlorine gas is approximately 5964.4 mL.
