What is the sum of the coefficients for the following reaction: \( \mathrm{H} 2 \mathrm{SO}_{4}+\mathrm{Al}(\mathrm{OH}) 3 \)--> \( \mathrm{Al} 2(\mathrm{SO} 4) 3+\mathrm{H} 2 \mathrm{O} \)

To find the sum of the coefficients in the balanced chemical equation, we first need to balance the equation. The unbalanced equation you provided is: \( \mathrm{H}_2\mathrm{SO}_4 + \mathrm{Al}(\mathrm{OH})_3 \rightarrow \mathrm{Al}_2(\mathrm{SO}_4)_3 + \mathrm{H}_2\mathrm{O} \) Let's balance it step by step: 1. Balance the aluminum (Al) atoms: There are 1 Al atom on the left and 2 Al atoms on the right. To balance the Al, we need to put a coefficient of 2 in front of \( \mathrm{Al}(\mathrm{OH})_3 \): \( \mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{Al}(\mathrm{OH})_3 \rightarrow \mathrm{Al}_2(\mathrm{SO}_4)_3 + \mathrm{H}_2\mathrm{O} \) Now there are 2 Al atoms on both sides. 2. Balance the sulfur (S) atoms: There are 1 S atom on the left and 3 S atoms on the right. To balance the S, we need to put a coefficient of 3 in front of \( \mathrm{H}_2\mathrm{SO}_4 \): \( 3\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{Al}(\mathrm{OH})_3 \rightarrow \mathrm{Al}_2(\mathrm{SO}_4)_3 + \mathrm{H}_2\mathrm{O} \) Now there are 3 S atoms on both sides. 3. Balance the hydrogen (H) atoms: There are 6 H atoms on the left (from 3 \( \mathrm{H}_2\mathrm{SO}_4 \)) and 6 H atoms on the right (from 6 \( \mathrm{H}_2\mathrm{O} \)). 4. Balance the oxygen (O) atoms: There are 12 O atoms on the left (from 3 \( \mathrm{H}_2\mathrm{SO}_4 \)) and 12 O atoms on the right (from 3 \( \mathrm{Al}_2(\mathrm{SO}_4)_3 \) and 6 \( \mathrm{H}_2\mathrm{O} \)). The balanced equation is: \( 3\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{Al}(\mathrm{OH})_3 \rightarrow \mathrm{Al}_2(\mathrm{SO}_4)_3 + 6\mathrm{H}_2\mathrm{O} \) Now, to find the sum of the coefficients, we add them together: \( 3 + 2 + 1 + 6 = 12 \) The sum of the coefficients in the balanced equation is 12.