A solution containing 5.00g of solid Al2(CO3)3 dissolved in water to a total volume of 200. mL is combined with 140. mL of 1.0 M HBr.
Write the total of ionic equation for the reaction described below and find the molar concentration of all ions remaining in a solution that is prepared by mixing the compounds listed below. Assume all equeous products are saluble.
I should point out that Al2(CO3)3 is not soluble in water.
Convert 5.00 g Al2(CO3)2 to mols. mol = grams/molar mass.
Convert HBr to mols mols = M x L = ?
Use the equation below to determine the limiting reagent and use that information to determine how much of each reagent remains. Then you can write the ionic equation.
Al2(CO3)3 + 6HBr ==> 2AlBr3 + 3H2O + 3CO2
To write the total ionic equation for the reaction and calculate the molar concentrations of the remaining ions, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction.
First, we need to write the balanced chemical equation for the reaction between Al2(CO3)3 and HBr:
Al2(CO3)3 + 6HBr → 2AlBr3 + 3H2CO3
Step 2: Break apart soluble compounds into their ions.
Next, we need to separate the compounds that are soluble into their respective ions. In this case, Al2(CO3)3 and H2CO3 are insoluble, so they should remain as whole compounds.
Al2(CO3)3 + 6H+ + 6Br- → 2Al3+ + 6Br- + 3H2CO3
Step 3: Cancel out spectator ions.
Spectator ions are those that do not participate in the reaction as they remain the same on both sides. In this case, the Br- ions are spectator ions on both sides and can be canceled out.
Al2(CO3)3 + 6H+ → 2Al3+ + 3H2CO3
Step 4: Calculate the molar concentrations of the remaining ions.
To find the molar concentrations of the remaining ions, we'll start with the given information:
- Initial mass of Al2(CO3)3: 5.00 g
- Total volume of the solution: 200 mL
- Volume of HBr added: 140 mL
- Molar concentration of HBr: 1.0 M
First, let's convert the volume of the solution and HBr to liters:
Total volume of the solution = 200 mL = 200/1000 = 0.200 L
Volume of HBr added = 140 mL = 140/1000 = 0.140 L
Next, we can calculate the number of moles of Al2(CO3)3:
Molar mass of Al2(CO3)3 = (2*26.98) + (3*12.01) + (9*16.00) = 381.61 g/mol
Number of moles of Al2(CO3)3 = mass/molar mass = 5.00 g / 381.61 g/mol = 0.0131 mol
Since each Al2(CO3)3 produces 2 moles of Al3+ ions in the reaction, the number of moles of Al3+ ions is:
Number of moles of Al3+ ions = 2 * (0.0131 mol) = 0.0262 mol
Similarly, since each HBr produces 6 moles of H+ ions in the reaction, the number of moles of H+ ions is:
Number of moles of H+ ions = 6 * (0.140 L * 1.0 M) = 0.840 mol
Finally, to find the molar concentration of each ion remaining in the solution, we divide the moles by the total volume of the solution:
Molar concentration of Al3+ ions = 0.0262 mol / 0.200 L = 0.131 M
Molar concentration of H+ ions = 0.840 mol / 0.200 L = 4.20 M
Therefore, the molar concentration of Al3+ ions remaining in the solution is 0.131 M, and the molar concentration of H+ ions remaining in the solution is 4.20 M.