If 840 mL of 0.0140 M aqueous Al3+ and 500 mL of 0.0680 M aqueous Br- are reacted, what mass (g) of Br- is produced?
Al2(SO4)3(aq) + 3 CaBr2(aq) → 2 AlBr3(aq) + 3 CaSO4(s)
Molar Mass (g)
Br- 79.904
Al3+ 26.982
Br- 79.904
Molar Volume (L)
22.400 at STP
Gas Constant
0.0821
To determine the mass of Br- produced, we need to calculate the limiting reagent in the reaction.
Step 1: Calculate the number of moles of Al3+ and Br- in their respective solutions.
- For Al3+:
- Volume of Al3+ solution = 840 mL = 0.840 L
- Concentration of Al3+ solution = 0.0140 M
- Number of moles of Al3+ = concentration * volume = 0.0140 * 0.840 = 0.0118 moles
- For Br-:
- Volume of Br- solution = 500 mL = 0.500 L
- Concentration of Br- solution = 0.0680 M
- Number of moles of Br- = concentration * volume = 0.0680 * 0.500 = 0.034 moles
Step 2: Calculate the stoichiometric ratio between Al3+ and Br- in the balanced chemical equation.
- From the balanced equation, we can see that the stoichiometric ratio between Al3+ and Br- is 2:3. This means that for every 2 moles of Al3+ reacted, 3 moles of Br- are required.
Step 3: Determine the limiting reagent.
- To determine the limiting reagent, we compare the number of moles of each reactant to the stoichiometric ratio.
- The number of moles of Al3+ is 0.0118 moles, and the number of moles of Br- is 0.034 moles.
- Using the stoichiometric ratio, we can calculate the expected number of moles of Br- required based on the number of moles of Al3+.
- For 0.0118 moles of Al3+, we would require (0.0118 moles * 3/2) = 0.0177 moles of Br-.
Since we have 0.034 moles of Br- available and we only need 0.0177 moles, the Br- is in excess and Al3+ is the limiting reagent.
Step 4: Calculate the mass of Br- produced.
- To calculate the mass of Br- produced, we need to know the molar mass of Br-.
- The molar mass of Br- is given as 79.904 g/mol.
- The number of moles of Br- produced is equal to the number of moles of Br- required, which is 0.0177 moles.
- To calculate the mass of Br-, we use the formula: mass = number of moles * molar mass
- The mass of Br- produced is (0.0177 moles * 79.904 g/mol) = 1.414 g.
Therefore, the mass of Br- produced in the reaction is 1.414 grams.