Problem
Given the balanced equation representing a reaction: \[ \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+6 \mathrm{NaOH} \rightarrow 2 \mathrm{Al}(\mathrm{OH})_{3}+3 \mathrm{Na}_{2} \mathrm{SO}_{4} \] The mole ratio of \( \mathrm{NaOH} \) to \( \mathrm{Al}(\mathrm{OH})_{3} \) is 1. \( 1: 1 \) 2. \( 1: 3 \) 3. \( 3: 1 \) 4. \( 3: 7 \)
Solution
Step 1 ：Identify the coefficients of the reactants and products from the balanced equation: \(6\mathrm{NaOH} \rightarrow 2\mathrm{Al}(\mathrm{OH})_{3}\)
Step 2 ：Write the mole ratio: \(\frac{6\mathrm{NaOH}}{2\mathrm{Al}(\mathrm{OH})_{3}}\)
Step 3 ：Reduce the fraction to lowest terms: \(\frac{3\mathrm{NaOH}}{1\mathrm{Al}(\mathrm{OH})_{3}}\)
