Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
To solve this problem, let's carefully go through the given reaction and the quantities involved step by step.
1. Identify the Molar Masses:
- The molar mass of [tex]\( Al_2(SO_4)_3 \)[/tex] is given as [tex]\( 342.15 \, \text{g/mol} \)[/tex].
- The molar mass of [tex]\( HCl \)[/tex] is given as [tex]\( 36.46 \, \text{g/mol} \)[/tex].
2. Convert the Volume of [tex]\( HCl \)[/tex] to Liters:
- The volume of [tex]\( HCl \)[/tex] solution is [tex]\( 500 \, \text{mL} \)[/tex], which is equivalent to [tex]\( 0.500 \, \text{L} \)[/tex].
3. Calculate the Moles of [tex]\( Al_2(SO_4)_3 \)[/tex]:
- Given mass of [tex]\( Al_2(SO_4)_3 \)[/tex] is [tex]\( 76.3 \, \text{g} \)[/tex].
- Moles of [tex]\( Al_2(SO_4)_3 \)[/tex] = [tex]\(\frac{\text{mass}}{\text{molar mass}} = \frac{76.3 \, \text{g}}{342.15 \, \text{g/mol}} = 0.223 \, \text{mol}\)[/tex].
4. Calculate the Moles of [tex]\( HCl \)[/tex]:
- Molarity of [tex]\( HCl \)[/tex] is [tex]\( 3.16 \, \text{M} \)[/tex].
- Volume of [tex]\( HCl \)[/tex] is [tex]\( 0.500 \, \text{L} \)[/tex].
- Moles of [tex]\( HCl \)[/tex] = [tex]\(\text{molarity} \times \text{volume} = 3.16 \, \text{mol/L} \times 0.500 \, \text{L} = 1.58 \, \text{mol}\)[/tex].
5. Relate the Reactants via Stoichiometry:
- The balanced reaction shows that [tex]\( 1 \)[/tex] mole of [tex]\( Al_2(SO_4)_3 \)[/tex] reacts with [tex]\( 6 \)[/tex] moles of [tex]\( HCl \)[/tex] to produce [tex]\( 3 \)[/tex] moles of [tex]\( H_2SO_4 \)[/tex].
- Therefore, moles of [tex]\( HCl \)[/tex] needed = [tex]\( \text{moles of } Al_2(SO_4)_3 \times 6 = 0.223 \times 6 = 1.34 \, \text{mol} \)[/tex].
6. Calculate the Moles of [tex]\( H_2SO_4 \)[/tex] Produced:
- From the stoichiometric relation, [tex]\( 1 \)[/tex] mole of [tex]\( Al_2(SO_4)_3 \)[/tex] produces [tex]\( 3 \)[/tex] moles of [tex]\( H_2SO_4 \)[/tex].
- Moles of [tex]\( H_2SO_4 \)[/tex] produced = [tex]\( \text{moles of } Al_2(SO_4)_3 \times 3 = 0.223 \times 3 = 0.669 \, \text{mol} \)[/tex].
7. Calculate the Final Molarity of [tex]\( H_2SO_4 \)[/tex]:
- Assume the volume of the solution remains unchanged at [tex]\( 0.500 \, \text{L} \)[/tex].
- Final molarity of [tex]\( H_2SO_4 \)[/tex] = [tex]\(\frac{\text{moles of } H_2SO_4}{\text{volume of solution}} = \frac{0.669 \, \text{mol}}{0.500 \, \text{L}} = 1.338 \, \text{M}\)[/tex].
Therefore, the final molarity of [tex]\( H_2SO_4 \)[/tex] is [tex]\( 1.338 \, \text{M} \)[/tex].
1. Identify the Molar Masses:
- The molar mass of [tex]\( Al_2(SO_4)_3 \)[/tex] is given as [tex]\( 342.15 \, \text{g/mol} \)[/tex].
- The molar mass of [tex]\( HCl \)[/tex] is given as [tex]\( 36.46 \, \text{g/mol} \)[/tex].
2. Convert the Volume of [tex]\( HCl \)[/tex] to Liters:
- The volume of [tex]\( HCl \)[/tex] solution is [tex]\( 500 \, \text{mL} \)[/tex], which is equivalent to [tex]\( 0.500 \, \text{L} \)[/tex].
3. Calculate the Moles of [tex]\( Al_2(SO_4)_3 \)[/tex]:
- Given mass of [tex]\( Al_2(SO_4)_3 \)[/tex] is [tex]\( 76.3 \, \text{g} \)[/tex].
- Moles of [tex]\( Al_2(SO_4)_3 \)[/tex] = [tex]\(\frac{\text{mass}}{\text{molar mass}} = \frac{76.3 \, \text{g}}{342.15 \, \text{g/mol}} = 0.223 \, \text{mol}\)[/tex].
4. Calculate the Moles of [tex]\( HCl \)[/tex]:
- Molarity of [tex]\( HCl \)[/tex] is [tex]\( 3.16 \, \text{M} \)[/tex].
- Volume of [tex]\( HCl \)[/tex] is [tex]\( 0.500 \, \text{L} \)[/tex].
- Moles of [tex]\( HCl \)[/tex] = [tex]\(\text{molarity} \times \text{volume} = 3.16 \, \text{mol/L} \times 0.500 \, \text{L} = 1.58 \, \text{mol}\)[/tex].
5. Relate the Reactants via Stoichiometry:
- The balanced reaction shows that [tex]\( 1 \)[/tex] mole of [tex]\( Al_2(SO_4)_3 \)[/tex] reacts with [tex]\( 6 \)[/tex] moles of [tex]\( HCl \)[/tex] to produce [tex]\( 3 \)[/tex] moles of [tex]\( H_2SO_4 \)[/tex].
- Therefore, moles of [tex]\( HCl \)[/tex] needed = [tex]\( \text{moles of } Al_2(SO_4)_3 \times 6 = 0.223 \times 6 = 1.34 \, \text{mol} \)[/tex].
6. Calculate the Moles of [tex]\( H_2SO_4 \)[/tex] Produced:
- From the stoichiometric relation, [tex]\( 1 \)[/tex] mole of [tex]\( Al_2(SO_4)_3 \)[/tex] produces [tex]\( 3 \)[/tex] moles of [tex]\( H_2SO_4 \)[/tex].
- Moles of [tex]\( H_2SO_4 \)[/tex] produced = [tex]\( \text{moles of } Al_2(SO_4)_3 \times 3 = 0.223 \times 3 = 0.669 \, \text{mol} \)[/tex].
7. Calculate the Final Molarity of [tex]\( H_2SO_4 \)[/tex]:
- Assume the volume of the solution remains unchanged at [tex]\( 0.500 \, \text{L} \)[/tex].
- Final molarity of [tex]\( H_2SO_4 \)[/tex] = [tex]\(\frac{\text{moles of } H_2SO_4}{\text{volume of solution}} = \frac{0.669 \, \text{mol}}{0.500 \, \text{L}} = 1.338 \, \text{M}\)[/tex].
Therefore, the final molarity of [tex]\( H_2SO_4 \)[/tex] is [tex]\( 1.338 \, \text{M} \)[/tex].
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
