To determine how many moles of [tex]\( SF_6 \)[/tex] can form from the combination of 11 moles of [tex]\( S \)[/tex] and 18 moles of [tex]\( F_2 \)[/tex], we need to analyze the balanced chemical equation:
[tex]\[
S + 3F_2 \rightarrow SF_6
\][/tex]
This equation tells us that 1 mole of sulfur ([tex]\( S \)[/tex]) reacts with 3 moles of fluorine ([tex]\( F_2 \)[/tex]) to produce 1 mole of sulfur hexafluoride ([tex]\( SF_6 \)[/tex]).
We first need to identify the limiting reagent, which is the reactant that will be completely consumed and thus limit the amount of product formed.
1. Determine how many moles of [tex]\( F_2 \)[/tex] are needed to react with 11 moles of [tex]\( S \)[/tex]:
[tex]\[
\text{Required moles of } F_2 = 11 \text{ moles of } S \times 3 \text{ moles of } F_2 \text{ per mole of } S = 33 \text{ moles of } F_2
\][/tex]
We only have 18 moles of [tex]\( F_2 \)[/tex], which is less than the 33 moles required. Therefore, [tex]\( F_2 \)[/tex] is the limiting reagent.
2. Determine how many moles of [tex]\( S \)[/tex] can react with 18 moles of [tex]\( F_2 \)[/tex]:
[tex]\[
\text{Reacting moles of } S = \frac{18 \text{ moles of } F_2}{3 \text{ moles of } F_2 \text{ per mole of } S} = 6 \text{ moles of } S
\][/tex]
Given that [tex]\( F_2 \)[/tex] is the limiting reagent and it will be completely consumed, only 6 moles of [tex]\( S \)[/tex] will react.
3. Determine the moles of [tex]\( SF_6 \)[/tex] that can be formed:
Since 6 moles of [tex]\( S \)[/tex] react with 18 moles of [tex]\( F_2 \)[/tex] to form 6 moles of [tex]\( SF_6 \)[/tex]:
[tex]\[
6 \text{ moles of } S \text{ produce } 6 \text{ moles of } SF_6
\][/tex]
Hence, the correct answer is:
[tex]\[
6 \text{ moles of } SF_6
\][/tex]
