Calculate $CO_2$ from Butane Oxidation

This solution details the calculation for the mass of $CO_2$ produced from the complete oxidation of $150 \text{ g}$ of butane ($C_4H_{10}$).

Step 1: Balanced Chemical Equation

The complete oxidation of butane ($C_4H_{10}$) involves reacting it with oxygen ($O_2$) to produce carbon dioxide ($CO_2$) and water ($H_2O$). The balanced chemical equation is essential for stoichiometric calculations:

$ 2 C_4H_{10} + 25 O_2 \rightarrow 8 CO_2 + 10 H_2O $

This equation shows that 2 moles of butane produce 8 moles of $CO_2$.

Step 2: Molar Masses Calculation

To relate mass to moles, we need the molar masses of the relevant compounds:

• Molar mass of butane ($C_4H_{10}$): $ (4 \times \text{Atomic Mass of C}) + (10 \times \text{Atomic Mass of H}) $ $ (4 \times 12.011 \text{ g/mol}) + (10 \times 1.008 \text{ g/mol}) \approx 58.12 \text{ g/mol} $
• Molar mass of carbon dioxide ($CO_2$): $ (1 \times \text{Atomic Mass of C}) + (2 \times \text{Atomic Mass of O}) $ $ (1 \times 12.011 \text{ g/mol}) + (2 \times 15.999 \text{ g/mol}) \approx 44.01 \text{ g/mol} $

Step 3: Stoichiometric Calculation

First, convert the given mass of butane to moles:

$ \text{Moles of } C_4H_{10} = \frac{\text{Mass of } C_4H_{10}}{\text{Molar Mass of } C_4H_{10}} = \frac{150 \text{ g}}{58.12 \text{ g/mol}} \approx 2.58 \text{ mol} $

Using the mole ratio from the balanced equation ($2 \text{ mol } C_4H_{10} : 8 \text{ mol } CO_2$, or $1:4$), calculate the moles of $CO_2$ produced:

$ \text{Moles of } CO_2 = 2.58 \text{ mol } C_4H_{10} \times \frac{8 \text{ mol } CO_2}{2 \text{ mol } C_4H_{10}} = 2.58 \times 4 \approx 10.32 \text{ mol } CO_2 $

Finally, convert the moles of $CO_2$ back to mass:

$ \text{Mass of } CO_2 = \text{Moles of } CO_2 \times \text{Molar Mass of } CO_2 $

$ \text{Mass of } CO_2 = 10.32 \text{ mol} \times 44.01 \text{ g/mol} \approx 454.15 \text{ g} $

Step 4: Final Answer Determination

The calculated mass of $CO_2$ is approximately $454.15 \text{ g}$. Rounding this value to the nearest whole number gives $455 \text{ g}$.