What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200M Al(No3)3 is added to 200 Ml of a 0.100M KoH
Balance the equation.
Find which reactant is in excess.
Then, for the limiting reactant, calculate the product moles.
If you have difficulty, I will be happy to critique your work.
.520
To determine the mass of solid aluminum hydroxide (Al(OH)3) that can be produced when 50.0 mL of 0.200 M Al(NO3)3 is added to 200 mL of a 0.100 M KOH solution, we need to calculate the limiting reagent and then use stoichiometry to find the mass.
Step 1: Calculate the number of moles of Al(NO3)3 and KOH:
Given:
Volume of Al(NO3)3 = 50.0 mL
Concentration of Al(NO3)3 = 0.200 M
Volume of KOH = 200 mL
Concentration of KOH = 0.100 M
To find moles of Al(NO3)3:
moles = concentration * volume (in liters)
moles of Al(NO3)3 = 0.200 M * (50.0 mL / 1000 mL/L) = 0.0100 moles
To find moles of KOH:
moles of KOH = 0.100 M * (200 mL / 1000 mL/L) = 0.0200 moles
Step 2: Determine the stoichiometry of the balanced chemical equation:
The balanced chemical equation representing the reaction between Al(NO3)3 and KOH is:
Al(NO3)3 + 3KOH -> Al(OH)3 + 3KNO3
From the equation, we can see that for every 1 mole of Al(NO3)3, 1 mole of Al(OH)3 is produced.
Step 3: Calculate the limiting reagent:
The limiting reagent is the reactant that is consumed completely and limits the amount of product that can be formed. In this case, we compare the moles of Al(NO3)3 and KOH.
Since the stoichiometry of the balanced equation is 1:1, we compare the moles of Al(NO3)3 and KOH in a 1:1 ratio.
moles of Al(NO3)3 : moles of KOH = 0.0100 : 0.0200 = 1 : 2
From the ratio, we can see that KOH is the limiting reagent because there are twice as many moles of KOH as Al(NO3)3. This means that all of the Al(NO3)3 will be consumed and used up in the reaction, while there will be excess KOH remaining.
Step 4: Calculate the moles of Al(OH)3 produced:
Since KOH is the limiting reagent, it determines the amount of Al(OH)3 produced. As per the balanced chemical equation, 1 mole of Al(OH)3 is produced for every 1 mole of Al(NO3)3.
moles of Al(OH)3 = moles of limiting reagent (KOH) = 0.0100 moles
Step 5: Calculate the mass of Al(OH)3 produced:
To calculate the mass of Al(OH)3, we need its molar mass. The molar mass of Al(OH)3 is the sum of the atomic masses of each element:
- Atomic mass of Al = 26.98 g/mol
- Atomic mass of O = 16.00 g/mol (3 O atoms)
- Atomic mass of H = 1.01 g/mol (3 H atoms)
Molar mass of Al(OH)3 = (26.98 g/mol) + (3 * 1.01 g/mol) + (3 * 16.00 g/mol) = 78.00 g/mol
Now we can calculate the mass of Al(OH)3 produced:
mass = moles * molar mass
mass of Al(OH)3 = 0.0100 moles * 78.00 g/mol = 0.780 g
Therefore, the mass of solid aluminum hydroxide produced is 0.780 g.