!A 100.0 mL aliquot of 0.200 M aqueous calcium hydroxide is mixed with ! 100.0 mL of 0.200 M aqueous aluminum nitrate.
a.)! Write a complete molecular equation for this reaction, including phases b.)! Write a net ionic equation for this reaction, including phases.
c.) !For each reacting ion, calculate the number of moles available.
d.) !Use the net ionic equation to determine the limiting reactant.
e.)! Calculate the mass of the precipitate produced.
f.)! Calculate the concentration for each ion remaining after the reaction is complete.
You can stop here.
a) The complete molecular equation for this reaction is:
Ca(OH)2(aq) + 2Al(NO3)3(aq) -> 2Al(OH)3(s) + 3Ca(NO3)2(aq)
b) The net ionic equation for this reaction is:
Ca2+(aq) + 2OH-(aq) + 2Al3+(aq) -> 2Al(OH)3(s) + Ca2+(aq)
c) To calculate the number of moles available for each reacting ion, we will use the formula:
moles = concentration * volume (in liters)
For calcium hydroxide (Ca(OH)2):
moles of Ca(OH)2 = 0.200 mol/L * 0.100 L = 0.020 mol
For aluminum nitrate (Al(NO3)3):
moles of Al(NO3)3 = 0.200 mol/L * 0.100 L = 0.020 mol
d) The net ionic equation tells us that 1 mole of calcium ions (Ca2+) reacts with 2 moles of hydroxide ions (OH-) and 2 moles of aluminum ions (Al3+) react with 2 moles of hydroxide ions (OH-). Therefore, the limiting reactant is the one that has the smallest ratio of moles to the coefficient in the balanced equation, which in this case is Ca2+.
e) The net ionic equation tells us that 1 mole of calcium ions (Ca2+) reacts with 1 mole of aluminum hydroxide (Al(OH)3) to form 1 mole of precipitate. Therefore, the mass of the precipitate produced can be calculated using the molar mass of aluminum hydroxide:
mass of precipitate = moles of Al(OH)3 * molar mass of Al(OH)3
f) After the reaction is complete, some ions will remain in solution. To calculate the concentration of each remaining ion, we need to consider the initial volumes and the moles of reactants consumed.
For calcium ions (Ca2+):
moles remaining = moles of Ca(NO3)2 - moles of Ca(OH)2 reacted
concentration remaining = moles remaining / total volume (in liters)
For aluminum ions (Al3+):
moles remaining = moles of Al(NO3)3 - (moles of Al(OH)3 formed * 2)
concentration remaining = moles remaining / total volume (in liters)
a.) To write the complete molecular equation, we need to identify the reactants and the product of the reaction. Given that we have a 100.0 mL aliquot of 0.200 M aqueous calcium hydroxide and 100.0 mL of 0.200 M aqueous aluminum nitrate, the reactants are calcium hydroxide (Ca(OH)2) and aluminum nitrate (Al(NO3)3). The molecular equation is:
Ca(OH)2(aq) + Al(NO3)3(aq) → Ca(NO3)2(aq) + Al(OH)3(s)
In this equation, (aq) represents an aqueous solution, and (s) represents a solid precipitate.
b.) To write the net ionic equation, we need to eliminate any spectator ions that do not participate in the actual reaction. The complete ionic equation for the reaction is:
Ca2+(aq) + 2OH-(aq) + Al3+(aq) + 3NO3-(aq) → Ca2+(aq) + 2NO3-(aq) + Al(OH)3(s)
To obtain the net ionic equation, remove the spectator ions:
2OH-(aq) + Al3+(aq) → Al(OH)3(s)
c.) To calculate the moles of each reacting ion, we need to determine the number of moles using the given concentration and volume. We know that the volume is 100.0 mL for both solutions, and the concentration is 0.200 M.
Number of moles of Ca2+ = 0.2 mol/L x 0.1 L = 0.02 moles
Number of moles of OH- = 0.2 mol/L x 0.1 L = 0.02 moles
Number of moles of Al3+ = 0.2 mol/L x 0.1 L = 0.02 moles
Number of moles of NO3- = 0.2 mol/L x 0.1 L = 0.02 moles
d.) To determine the limiting reactant, we compare the moles of each reacting ion. The reaction ratio is 1:1 for Al3+ to OH-. Since both reactants have equal moles (0.02 moles), neither is limiting.
e.) To calculate the mass of the precipitate produced, we need to use stoichiometry. From the net ionic equation, we can see that 1 mole of Al(OH)3 is formed for every 1 mole of Al3+. Therefore, the mass of Al(OH)3 can be calculated using its molar mass.
The molar mass of Al(OH)3 = 78.0 g/mol
Mass of Al(OH)3 = Moles of Al(OH)3 x Molar mass of Al(OH)3
= 0.02 moles x 78.0 g/mol
= 1.56 g
f.) To calculate the concentration of each ion remaining after the reaction is complete, we need to consider the stoichiometry of the reaction and the volume of the final solution. Since the reaction is complete, all the Ca2+ and NO3- ions will remain in solution, while the OH- ions will react with Al3+ to form Al(OH)3.
Concentration of Ca2+ = 0.200 M (unchanged)
Concentration of OH- = 0 M (reacted completely)
Concentration of Al3+ = 0 M (reacted completely)
Concentration of NO3- = 0.200 M (unchanged)
Note: The moles of OH- and Al3+ ions are zero because they react completely, leaving none in the final solution.