Given the balanced equation representing a reaction:
${Al}_{2} {({SO}_{4})}_{3} + 6 NaOH \to 2 Al (OH)_{3} + 3 {Na}_{2} {SO}_{4}$
The mole ratio of $NaOH$ to $Al (OH)_{3}$ is
1. $1 : 1$
2. $1 : 3$
3. $3 : 1$
4. $3 : 7$

Answer

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Answer

Reduce the fraction to lowest terms: $\frac{3 NaOH}{1 Al (OH)_{3}}$

Steps

Step 1 ：Identify the coefficients of the reactants and products from the balanced equation: $6 NaOH \to 2 Al (OH)_{3}$

Step 2 ：Write the mole ratio: $\frac{6 NaOH}{2 Al (OH)_{3}}$

Step 3 ：Reduce the fraction to lowest terms: $\frac{3 NaOH}{1 Al (OH)_{3}}$