Let's analyze the given chemical reaction:
[tex]\[ \text{FeCl}_2 + 2\ \text{NaOH} \rightarrow \text{Fe(OH)}_2 (s) + 2\ \text{NaCl} \][/tex]
Initially, we are given:
- 6 moles of [tex]\( \text{FeCl}_2 \)[/tex]
- 6 moles of [tex]\( \text{NaOH} \)[/tex]
The balanced equation tells us that:
- 1 mole of [tex]\( \text{FeCl}_2 \)[/tex] reacts with 2 moles of [tex]\( \text{NaOH} \)[/tex]
First, we need to determine the stoichiometric requirements for a complete reaction. For every 1 mole of [tex]\( \text{FeCl}_2 \)[/tex], 2 moles of [tex]\( \text{NaOH} \)[/tex] are needed.
Given:
- 6 moles of [tex]\( \text{FeCl}_2 \)[/tex] will need [tex]\( 6 \times 2 = 12 \)[/tex] moles of [tex]\( \text{NaOH} \)[/tex]
However, we only have 6 moles of [tex]\( \text{NaOH} \)[/tex], which is insufficient for 6 moles of [tex]\( \text{FeCl}_2 \)[/tex]. Therefore, [tex]\( \text{NaOH} \)[/tex] is the limiting reactant.
We can now determine how many moles of [tex]\( \text{FeCl}_2 \)[/tex] can react with the available 6 moles of [tex]\( \text{NaOH} \)[/tex]:
- Since 2 moles of [tex]\( \text{NaOH} \)[/tex] are needed for every 1 mole of [tex]\( \text{FeCl}_2 \)[/tex], the 6 moles of [tex]\( \text{NaOH} \)[/tex] can react with [tex]\( \frac{6}{2} = 3 \)[/tex] moles of [tex]\( \text{FeCl}_2 \)[/tex].
Thus, 3 moles of [tex]\( \text{FeCl}_2 \)[/tex] will be used up in the reaction.
So, the correct answer is:
C. 3
