To determine how many grams of [tex]\( \text{Al(OH)}_3 \)[/tex] are required to completely react with 87.061 g of [tex]\( \text{H}_2\text{SO}_4 \)[/tex], we need to follow these steps:
1. Write the balanced chemical equation:
[tex]\[
2 \text{Al(OH)}_3 + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6 \text{H}_2\text{O}
\][/tex]
From the balanced equation, we see that 2 moles of [tex]\( \text{Al(OH)}_3 \)[/tex] react with 3 moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex].
2. Calculate the molar masses:
- The molar mass of [tex]\( \text{Al(OH)}_3 \)[/tex] is approximately 78.0 g/mol.
- The molar mass of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] is approximately 98.079 g/mol.
3. Determine the moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] present:
Given mass of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] is 87.061 g.
[tex]\[
\text{Moles of } \text{H}_2\text{SO}_4 = \frac{\text{mass of } \text{H}_2\text{SO}_4}{\text{molar mass of } \text{H}_2\text{SO}_4} = \frac{87.061 \text{ g}}{98.079 \text{ g/mol}} \approx 0.8877 \text{ moles}
\][/tex]
4. Use stoichiometry to find the moles of [tex]\( \text{Al(OH)}_3 \)[/tex] needed:
According to the balanced chemical equation, 3 moles of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] react with 2 moles of [tex]\( \text{Al(OH)}_3 \)[/tex]:
[tex]\[
\text{Moles of } \text{Al(OH)}_3 \text{ needed} = \left(\frac{2}{3}\right) \times \text{moles of } \text{H}_2\text{SO}_4 = \left(\frac{2}{3}\right) \times 0.8877 \approx 0.5918 \text{ moles}
\][/tex]
5. Calculate the mass of [tex]\( \text{Al(OH)}_3 \)[/tex] needed:
[tex]\[
\text{Mass of } \text{Al(OH)}_3 = \text{moles of } \text{Al(OH)}_3 \times \text{molar mass of } \text{Al(OH)}_3 = 0.5918 \text{ moles} \times 78.0 \text{ g/mol} \approx 46.1584 \text{ g}
\][/tex]
Therefore, the mass of [tex]\( \text{Al(OH)}_3 \)[/tex] required to completely react with 87.061 g of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] is approximately 46.1584 grams.
