A 250.0 mL sample of 0.200 M aqueous aluminum chloride is mixed with 100.0 mL of 0.200 M aqueous sodium hydroxide.
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To determine what happens when aqueous aluminum chloride and sodium hydroxide are mixed together, we need to write out the balanced chemical equation.
The balanced equation for the reaction between aluminum chloride and sodium hydroxide is:
2AlCl3(aq) + 6NaOH(aq) → 2Al(OH)3(s) + 6NaCl(aq)
From the equation, we can see that two moles of aluminum chloride react with six moles of sodium hydroxide to produce two moles of aluminum hydroxide and six moles of sodium chloride.
To find the moles of each reactant, we can use the formula:
moles = concentration (M) × volume (L)
For the aluminum chloride solution, the moles can be calculated as follows:
moles of AlCl3 = 0.200 M × 0.250 L = 0.05 moles
For the sodium hydroxide solution, the moles can be calculated as follows:
moles of NaOH = 0.200 M × 0.100 L = 0.02 moles
From the balanced equation, we can see that the stoichiometric ratio between aluminum chloride and sodium hydroxide is 2:6. Therefore, we need to determine the limiting reactant to find out which reactant will be completely consumed and which one will be in excess.
To find the limiting reactant, we compare the moles of each reactant. In this case, the sodium hydroxide has the lowest moles (0.02 moles) compared to the aluminum chloride (0.05 moles). Therefore, sodium hydroxide is the limiting reactant.
Since sodium hydroxide is the limiting reactant, it will be completely consumed in the reaction. The remaining aluminum chloride will be in excess, and some of it will not participate in the reaction.
To find the amount of aluminum hydroxide and sodium chloride formed, we use the stoichiometry of the balanced chemical equation. From the balanced equation, we know that for every two moles of aluminum chloride, we get two moles of aluminum hydroxide. So, in this case, the moles of aluminum hydroxide formed would be equal to the moles of sodium hydroxide used.
Therefore, the moles of Al(OH)3 formed = 0.02 moles
To find the final concentration of the aluminum hydroxide solution, we divide the moles of Al(OH)3 by the total volume of the solution, which is the sum of the volumes of the reactants. The final volume of the solution is 250.0 mL + 100.0 mL = 350.0 mL = 0.350 L.
Concentration of Al(OH)3 = moles of Al(OH)3 / volume of solution
= 0.02 moles / 0.350 L
≈ 0.057 M
Therefore, the final concentration of the aluminum hydroxide solution is approximately 0.057 M.