To determine the mass of [tex]\( C_4H_{10} \)[/tex] required to produce 14.7 grams of [tex]\( CO_2 \)[/tex], we will follow a step-by-step approach:
1. Calculate the moles of [tex]\( CO_2 \)[/tex] produced:
We start by determining the number of moles of [tex]\( CO_2 \)[/tex] given the mass of [tex]\( CO_2 \)[/tex].
- The molar mass of [tex]\( CO_2 \)[/tex] is 44.01 g/mol.
[tex]\[
\text{Moles of } CO_2 = \frac{\text{mass of } CO_2}{\text{molar mass of } CO_2} = \frac{14.7 \text{ g}}{44.01 \text{ g/mol}} \approx 0.334 \text{ moles}
\][/tex]
2. Use stoichiometry to relate moles of [tex]\( CO_2 \)[/tex] to moles of [tex]\( C_4H_{10} \)[/tex]:
From the balanced chemical equation:
[tex]\[
2 C_4H_{10} + 13 O_2 \rightarrow 8 CO_2 + 10 H_2O
\][/tex]
We see that 2 moles of [tex]\( C_4H_{10} \)[/tex] produce 8 moles of [tex]\( CO_2 \)[/tex]. Using this ratio:
[tex]\[
\text{Moles of } C_4H_{10} = \left(\frac{2 \text{ moles of } C_4H_{10}}{8 \text{ moles of } CO_2}\right) \times 0.334 \text{ moles of } CO_2 \approx 0.084 \text{ moles}
\][/tex]
3. Convert moles of [tex]\( C_4H_{10} \)[/tex] to mass:
Finally, we convert the moles of [tex]\( C_4H_{10} \)[/tex] to mass in grams.
- The molar mass of [tex]\( C_4H_{10} \)[/tex] is 58.12 g/mol.
[tex]\[
\text{Mass of } C_4H_{10} = \text{moles of } C_4H_{10} \times \text{molar mass of } C_4H_{10} = 0.084 \text{ moles} \times 58.12 \text{ g/mol} \approx 4.853 \text{ grams}
\][/tex]
Thus, the mass of [tex]\( C_4H_{10} \)[/tex] required to produce 14.7 grams of [tex]\( CO_2 \)[/tex] is [tex]\( \boxed{4.853} \text{ grams} \)[/tex].
