Questions: 8. Balance the following reaction: Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O a. 4 Al(OH)3 + 6 H2SO4 → 2 Al2(SO4)3 + 12 H2O b. 3 Al(OH)3 + 2 H2SO4 → 6 Al2(SO4)3 + H2O c. 2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O d. 3 Al(OH)3 + 2 H2SO4 → 6 Al2(SO4)3 + H2O 9. Using the equation in question 8, if 0.25 g of Al(OH)3 is reacted how many grams of Al2(SO4)3 is produced? a. 1.1 grams b. 0.548 grams c. 0.605 grams d. 0.55 grams

8. Balance the following reaction:
Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O
a. 4 Al(OH)3 + 6 H2SO4 → 2 Al2(SO4)3 + 12 H2O
b. 3 Al(OH)3 + 2 H2SO4 → 6 Al2(SO4)3 + H2O
c. 2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O
d. 3 Al(OH)3 + 2 H2SO4 → 6 Al2(SO4)3 + H2O

9. Using the equation in question 8, if 0.25 g of Al(OH)3 is reacted how many grams of Al2(SO4)3 is produced?
a. 1.1 grams
b. 0.548 grams
c. 0.605 grams
d. 0.55 grams
Transcript text: 8. Balance the following reaction: \[ \mathrm{Al}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2} \mathrm{O} \] a. $4 \mathrm{Al}(\mathrm{OH})_{3}+6 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+12 \mathrm{H}_{2} \mathrm{O}$ b. $3 \mathrm{Al}(\mathrm{OH})_{3}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 6 \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2} \mathrm{O}$ c. $2 \mathrm{Al}(\mathrm{OH})_{3}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+6 \mathrm{H}_{2} \mathrm{O}$ d. $3 \mathrm{Al}(\mathrm{OH})_{3}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 6 \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{H}_{2} \mathrm{O}$ 9. Using the equation in question 8 , if 0.25 g of $\mathrm{Al}(\mathrm{OH})_{3}$ is reacted how many grams of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is produced? a. 1.1 grams b. 0.548 grams c. 0.605 grams d. 0.55 grams
failed

Solution

failed
failed

Solution Steps

Step 1: Identify the Correct Balanced Equation

We need to balance the given chemical reaction: \[ \mathrm{Al}(\mathrm{OH})_{3} + \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} + \mathrm{H}_{2}\mathrm{O} \]

By comparing the options provided:

  • Option (a) is balanced correctly: \[ 4 \mathrm{Al}(\mathrm{OH})_{3} + 6 \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow 2 \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} + 12 \mathrm{H}_{2}\mathrm{O} \]
Step 2: Calculate Molar Masses

Calculate the molar masses of the reactants and products:

  • Molar mass of \(\mathrm{Al}(\mathrm{OH})_{3}\): \[ \mathrm{Al}(\mathrm{OH})_{3} = 26.98 + 3(15.9994 + 1.0079) = 78.0036 \, \text{g/mol} \]
  • Molar mass of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\): \[ \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} = 2(26.98) + 3(32.065 + 4(15.9994)) = 342.1508 \, \text{g/mol} \]
Step 3: Use Stoichiometry to Find the Mass of Product

Using the balanced equation: \[ 4 \mathrm{Al}(\mathrm{OH})_{3} + 6 \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow 2 \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} + 12 \mathrm{H}_{2}\mathrm{O} \]

From the balanced equation, 4 moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) produce 2 moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\).

First, find the moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) in 0.25 g: \[ \text{Moles of } \mathrm{Al}(\mathrm{OH})_{3} = \frac{0.25 \, \text{g}}{78.0036 \, \text{g/mol}} = 0.0032 \, \text{mol} \]

Using the stoichiometric ratio: \[ \text{Moles of } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} = \frac{0.0032 \, \text{mol} \times 2}{4} = 0.0016 \, \text{mol} \]

Finally, convert moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) to grams: \[ \text{Mass of } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} = 0.0016 \, \text{mol} \times 342.1508 \, \text{g/mol} = 0.5474 \, \text{g} \]

Final Answer

\(\boxed{0.548 \, \text{grams}}\)

The answer is b. 0.548 grams.

Was this solution helpful?
failed
Unhelpful
failed
Helpful