To determine the limiting reactant between aluminum hydroxide (\( \text{Al(OH)}_3 \)) and sulfuric acid (\( \text{H}_2\text{SO}_4 \)), we first need the balanced chemical reaction for the reaction between these two compounds. The reaction can be given as follows:
\[ \text{Al(OH)}_3 + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6 \text{H}_2\text{O} \]
From the balanced equation, we can see that 1 mole of \( \text{Al(OH)}_3 \) reacts with 3 moles of \( \text{H}_2\text{SO}_4 \).
Next, we will calculate how many moles of \( \text{H}_2\text{SO}_4 \) are needed to completely react with the given 0.400 moles of \( \text{Al(OH)}_3 \):
\[ 0.400 , \text{mol Al(OH)}_3 \times \frac{3 , \text{mol H}_2\text{SO}_4}{1 , \text{mol Al(OH)}_3} = 1.200 , \text{mol H}_2\text{SO}_4 \]
We only have 0.400 moles of \( \text{H}_2\text{SO}_4 \) available, which is significantly less than the 1.200 moles required for complete reaction with \( \text{Al(OH)}_3 \).
Since we do not have enough \( \text{H}_2\text{SO}_4 \) to fully react with \( \text{Al(OH)}_3 \), the limiting reactant in this case is \( \text{H}_2\text{SO}_4 \).