Two solutions were prepared. The first was a 0.672 M solution of aluminum hydroxide and the second was a 0.405 M sulfuric acid solution.
1.) If a total of 500 mL of aluminum hydroxide solution was prepared, how many grams of the solute (AL(OH)3) would have been weighed out in order to make this solution.
2.) how many mL of the aluminum hydroxide solution would be used up if it was reacted with 1182 mL of sulfuric acid solution? How many mL of AL(OH)3 solution remains?
3.) How many grams of each of the products are formed as a result of this chemical reaction?
98.08 g/mol 78.00 g/mol
+
H2SO4 AL(OH)3
-------------------------->
342.15 g/mol 18.02 g/mol
+
AL2(SO4)3 H2O
2Al(OH)3 + 3H2SO4 --> Al2(SO4)3 + 6H2O
1) How many mols Al(OH)3 do you want? That's M x L = 0.672 x 0.500 = approx 0.3 but you should be more accurate than that.
mols Al(OH)3 = grams/molar mass. You know mols and molar mass; solve for grams.
2)How many mols H2SO4 did you add? That's M x L = 0.405 x 1.182 = approx 0.5
Convert mols H2SO4 to mols Al(OH)3.
Approx 0.5 mol H2SO4 x (2 mols Al(OH)3/3 mol H2SO4) = approx 0.5 x 2/3 = about 0.3 mols Al(OH)3 used. Using M x L = mols, convert 0.26 mols Al(OH)3 to L and convert to mL Al(OH)3 used. Subtract from what you had initially to find mL remaining.
3. This is a limiting reagent problem (LR) and you know that because amounts are given for both reactants. You already know from the problem that the LR is H2SO4. Here is how you do the g H2O produced. I'll leave the Al2(SO4)3 for you.
Convert mols H2SO4 to mols H2O. That's
approx 0.5 mol H2SO4 x (6 mols H2O/3 mols H2SO4) = 0.5 x 6/3 = about 1 mol H2O.
g H2O = mols H2O x molar mass H2O.
To find the answers to these questions, we need to use the concept of molarity and perform calculations based on the given information.
1.) To find the number of grams of solute (Al(OH)3) in the 500 mL solution, we can use the formula:
moles = molarity × volume
Volume = 500 mL = 0.5 L (since 1 L = 1000 mL)
Molarity = 0.672 M
moles = 0.672 mol/L × 0.5 L = 0.336 mol
To convert moles to grams, we need to know the molar mass of Al(OH)3, which is:
Al: 26.98 g/mol
O: 16.00 g/mol
H: 1.01 g/mol (x3 because there are three hydrogens)
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 grams of Al(OH)3:
grams = moles × molar mass
= 0.336 mol × 78.00 g/mol
= 26.208 g
Therefore, 26.208 grams of Al(OH)3 would have been weighed out to make the 500 mL solution.
2.) If 1182 mL of sulfuric acid solution reacts with the aluminum hydroxide solution, we need to determine the reaction stoichiometry to find out the ratio of reactants consumed.
The balanced equation for the reaction is:
2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O
From the equation, we can see that 2 moles of Al(OH)3 react with 3 moles of H2SO4.
Using the molarity and volume of the aluminum hydroxide solution, we can find the moles of Al(OH)3 in this solution:
moles of Al(OH)3 = molarity × volume
= 0.672 mol/L × 0.5 L
= 0.336 mol
From the reaction stoichiometry, we know that 2 moles of Al(OH)3 react with 3 moles of H2SO4. Therefore, we can find the moles of H2SO4 reacting:
moles of H2SO4 = (3/2) × moles of Al(OH)3
= (3/2) × 0.336 mol
= 0.504 mol
To find the volume of the aluminum hydroxide solution used, we need to use the molarity and the moles of Al(OH)3:
volume = moles / molarity
= 0.336 mol / 0.672 mol/L
= 0.5 L
Therefore, 500 mL of the aluminum hydroxide solution would be used up if it reacted with 1182 mL of the sulfuric acid solution.
To find the remaining volume of the aluminum hydroxide solution:
remaining volume = initial volume - volume used
= 500 mL - 500 mL
= 0 mL
Therefore, 0 mL of the aluminum hydroxide solution remains.
3.) To find the grams of each product formed, we need to consider the reaction stoichiometry.
From the balanced equation:
2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O
We can see that 2 moles of Al(OH)3 react to form 1 mole of Al2(SO4)3.
Using the moles of Al(OH)3 calculated earlier (0.336 mol), we can find the moles of Al2(SO4)3 formed:
moles of Al2(SO4)3 = (1/2) × moles of Al(OH)3
= (1/2) × 0.336 mol
= 0.168 mol
To convert moles to grams, we use the molar mass of Al2(SO4)3:
Molar mass of Al2(SO4)3 = (26.98 g/mol × 2) + (32.07 g/mol × 3 × 4) = 342.15 g/mol
Now we can calculate the grams of Al2(SO4)3 formed:
grams of Al2(SO4)3 = moles × molar mass
= 0.168 mol × 342.15 g/mol
= 57.52 g
From the balanced equation, we can see that 3 moles of H2SO4 react to form 6 moles of H2O.
Using the moles of H2SO4 calculated earlier (0.504 mol), we can find the moles of H2O formed:
moles of H2O = (6/3) × moles of H2SO4
= (6/3) × 0.504 mol
= 1.008 mol
To convert moles to grams, we use the molar mass of H2O:
Molar mass of H2O = 2.02 g/mol + 16.00 g/mol = 18.02 g/mol
Now we can calculate the grams of H2O formed:
grams of H2O = moles × molar mass
= 1.008 mol × 18.02 g/mol
= 18.18 g
Therefore, 57.52 grams of Al2(SO4)3 and 18.18 grams of H2O are formed as a result of this chemical reaction.