Determine the mass (in grams) of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$ that is required to produce $$ 14.7 \mathrm{~g} \mathrm{CO}_{2} $$. The balanced equation for the complete combustion of butane is shown below. $$ 2 \mathrm{C}_{4} \mathrm{H}_{10}+13 \mathrm{O}_{2} \rightarrow 8 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O} $$ - Your answer should have three significant figures.

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To determine the mass of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$ required to produce $$ 14.7 \, \text{g} \, \mathrm{CO}_{2} $$, we can follow these steps:

1. **Calculate the moles of $$ \mathrm{CO}_{2} $$** produced.
   The molar mass of $$ \mathrm{CO}_{2} $$ is calculated as follows:
   $$
   \text{Molar mass of } \mathrm{CO}_{2} = 12.01 \, \text{g/mol (C)} + 2 \times 16.00 \, \text{g/mol (O)} = 12.01 + 32.00 = 44.01 \, \text{g/mol}
   $$
   Now, we can calculate the moles of $$ \mathrm{CO}_{2} $$:
   $$
   \text{Moles of } \mathrm{CO}_{2} = \frac{14.7 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.334 \, \text{mol}
   $$

2. **Use the stoichiometry of the reaction** to find the moles of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$ needed.
   From the balanced equation:
   $$
   2 \mathrm{C}_{4} \mathrm{H}_{10} + 13 \mathrm{O}_{2} \rightarrow 8 \mathrm{CO}_{2} + 10 \mathrm{H}_{2} \mathrm{O}
   $$
   We see that 2 moles of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$ produce 8 moles of $$ \mathrm{CO}_{2} $$. Therefore, the ratio of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$ to $$ \mathrm{CO}_{2} $$ is:
   $$
   \frac{2 \, \text{mol } \mathrm{C}_{4} \mathrm{H}_{10}}{8 \, \text{mol } \mathrm{CO}_{2}} = \frac{1}{4}
   $$
   Thus, to find the moles of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$ needed:
   $$
   \text{Moles of } \mathrm{C}_{4} \mathrm{H}_{10} = \frac{1}{4} \times \text{Moles of } \mathrm{CO}_{2} = \frac{1}{4} \times 0.334 \, \text{mol} \approx 0.0835 \, \text{mol}
   $$

3. **Calculate the mass of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$** required.
   The molar mass of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$ is:
   $$
   \text{Molar mass of } \mathrm{C}_{4} \mathrm{H}_{10} = 4 \times 12.01 \, \text{g/mol (C)} + 10 \times 1.008 \, \text{g/mol (H)} = 48.04 + 10.08 = 58.12 \, \text{g/mol}
   $$
   Now, we can calculate the mass of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$:
   $$
   \text{Mass of } \mathrm{C}_{4} \mathrm{H}_{10} = \text{Moles of } \mathrm{C}_{4} \mathrm{H}_{10} \times \text{Molar mass of } \mathrm{C}_{4} \mathrm{H}_{10} = 0.0835 \, \text{mol} \times 58.12 \, \text{g/mol} \approx 4.86 \, \text{g}
   $$

4. **Round to three significant figures**:
   The final answer is:
   $$
   \boxed{4.86} \, \text{g}
   $$
To find the mass of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$ needed to produce $$ 14.7 \, \text{g} \, \mathrm{CO}_{2} $$, calculate the moles of $$ \mathrm{CO}_{2} $$ (0.334 mol), then use stoichiometry to find the moles of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$ (0.0835 mol), and finally multiply by the molar mass of $$ \mathrm{C}_{4} \mathrm{H}_{10} $$ (58.12 g/mol) to get the mass (4.86 g).

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