What mass of solid aluminum hydroxide can be produced when 50.0mL of .200 M Al(NO3)3 is added to 200.0 mL of .100 M KOH

0.520 g

Write the equation and balance it.

Al(NO3)3 + 3KOH ==> 3KNO3 + Al(OH)3
Calculate mols Al(NO3)3 by M x L = mols.
Calculate mols KOH by M x L = mols.
Using the equation, calculate how much Al(OH)3 can be produced by EACH reagent if you had all of the OTHER reagent you needed. The smaller of the two amounts (in mols) will be the amount produced.
Then mols x molar mass = g Al(OH)3.
Post your work if you get stuck.
Check my work.

To determine the mass of solid aluminum hydroxide produced, we need to use a balanced chemical equation that relates the reactants and products.

The balanced chemical equation for the reaction between aluminum nitrate (Al(NO3)3) and potassium hydroxide (KOH) is:

2 Al(NO3)3 + 6 KOH → 2 Al(OH)3 + 6 KNO3

From the balanced equation, we can see that 2 moles of Al(NO3)3 produce 2 moles of Al(OH)3.

Step 1: Calculate the moles of Al(NO3)3.
Moles of Al(NO3)3 = Volume (in L) × Concentration
Moles of Al(NO3)3 = 0.050 L × 0.200 M = 0.01 moles

Step 2: Use the mole ratio from the balanced equation to determine the moles of Al(OH)3.
Moles of Al(OH)3 = 0.01 moles

Step 3: Convert the moles of Al(OH)3 to grams using the molar mass of Al(OH)3.
Molar mass of Al(OH)3 = 78.00 g/mol
Mass of Al(OH)3 = Moles × Molar mass
Mass of Al(OH)3 = 0.01 moles × 78.00 g/mol = 0.78 g

Therefore, the mass of solid aluminum hydroxide that can be produced is 0.78 grams.

To determine the mass of solid aluminum hydroxide produced, you need to follow these steps:

Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction between aluminum nitrate (Al(NO3)3) and potassium hydroxide (KOH) to form aluminum hydroxide (Al(OH)3) is:

Al(NO3)3 + 3 KOH -> Al(OH)3 + 3 KNO3

Step 2: Determine the limiting reagent.
To determine the limiting reagent, compare the moles of each reactant:

For aluminum nitrate (Al(NO3)3):
Moles = (concentration in M) x (volume in L)
Moles of Al(NO3)3 = 0.200 M x (50.0 mL / 1000 mL/L) = 0.0100 moles

For potassium hydroxide (KOH):
Moles = (concentration in M) x (volume in L)
Moles of KOH = 0.100 M x (200.0 mL / 1000 mL/L) = 0.0200 moles

Based on the balanced equation, the stoichiometric ratio between Al(NO3)3 and KOH is 1:3. Therefore, 0.0100 moles of Al(NO3)3 will react with 0.0300 moles of KOH.

Since there is an excess of KOH (0.0200 moles), Al(NO3)3 is the limiting reagent.

Step 3: Calculate the moles of aluminum hydroxide formed.
From the balanced equation, the stoichiometric ratio between Al(NO3)3 and Al(OH)3 is 1:1. Therefore, the moles of Al(OH)3 formed will be equal to the moles of Al(NO3)3 used.

Moles of Al(OH)3 = Moles of Al(NO3)3 = 0.0100 moles

Step 4: Calculate the mass of aluminum hydroxide formed.
To calculate the mass of aluminum hydroxide formed, you need to use its molar mass.

The molar mass of Al(OH)3 = (atomic mass of Al) + 3 x (atomic mass of O) + 3 x (atomic mass of H)
= (26.98 g/mol) + 3 x (16.00 g/mol) + 3 x (1.01 g/mol)
= 78.00 g/mol

Mass of Al(OH)3 = Moles of Al(OH)3 x Molar mass of Al(OH)3
= 0.0100 moles x 78.00 g/mol
= 0.780 g

Therefore, 0.780 grams of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M Al(NO3)3 is added to 200.0 mL of 0.100 M KOH.