50.0 ml of 2.50M Sulfuric acid reacted with 43.5ml 2.75 Aluminum hydroxide. The reaction yielded Aluminum sulfate and water. (Al=27g/mol, S=32g/mol, H=1g/mol, O=16g/mol)
1. Write the balance chemical equation.
2. What is the limiting reactant?
3. How much reactant is in excess?
4. How many grams of Aluminum sulfate will be produced?
1.
3H2SO4 + 2Al(OH)3 ==> Al2(SO4)3 + 6H2O
2.
millimoles H2SO4 = mL x M = 50.0x 2.50 = 125
millimoles Al(OH)3 = 43.5 mL x 2.75 M = 119.625
Choose one; e.g. H2SO4. How much Al(OH)3 is required? That's
125 mmoles H2SO4 x (2 mol Al(OH)3/3 moles H2SO4) = 125 x 2/3 = 83.33 mmoles Al(OH)3. Do we have that much? Yes, so H2SO4 is the limiting reagent (LR) and Al(OH)3 is the excess reagent (ER). What if we chose the other one in our trial and error. That's 119.625 x (3 moles H2SO4/2 mols Al(OH)3) = 179.4 mmoles H2SO4 needed. We don't have the much H2SO4; therefore, H2SO4 is the LR and Al(OH)3 is the ER.
3.
The reaction will use all of the H2SO4 = 125 millmoles. We will use 83.33 millimoles Al(OH)3. We started with 119.625 millimoles Al(OH)3.The difference between 119.625 and 83.333 is the ER left unreacted.
4.
125 mmoles H2SO4 x (1 mole Al2(SO4)3/3 moles H2SO4) = 125 x 1/3 = 41.67 mmoles = 0.0416 mols. grams = mols x molar mass = ?
Check my work. CONFIRM ALL OF THE ABOVE.
1. To write the balanced chemical equation, we need to determine the formula for aluminum sulfate and water.
The formula for sulfuric acid is H2SO4, and aluminum hydroxide is Al(OH)3.
The balanced chemical equation for the reaction can be written as:
3H2SO4 + 2Al(OH)3 → Al2(SO4)3 + 6H2O
2. To determine the limiting reactant, we need to compare the moles of each reactant.
First, let's calculate the moles of sulfuric acid:
Moles of sulfuric acid = Volume of sulfuric acid (in L) × Concentration of sulfuric acid (in mol/L)
= 50.0 ml × (1 L/1000 ml) × 2.50 mol/L
= 0.125 mol
Next, let's calculate the moles of aluminum hydroxide:
Moles of aluminum hydroxide = Volume of aluminum hydroxide (in L) × Concentration of aluminum hydroxide (in mol/L)
= 43.5 ml × (1 L/1000 ml) × 2.75 mol/L
= 0.119625 mol
From the balanced equation, the stoichiometric ratio between sulfuric acid and aluminum hydroxide is 3:2. This means that theoretically, 3 moles of sulfuric acid react with 2 moles of aluminum hydroxide.
Comparing the moles in the reaction, it is clear that the amount of aluminum hydroxide is limiting. So, aluminum hydroxide is the limiting reactant.
3. To determine the amount of reactant in excess, we can subtract the amount of aluminum hydroxide consumed from the initial amount.
Moles of aluminum hydroxide in excess = Initial moles of aluminum hydroxide - Moles of aluminum hydroxide consumed
= 0.119625 mol - 0.119625 mol
= 0 mol
Since the moles of aluminum hydroxide consumed is equal to the initial moles of aluminum hydroxide, it means that there is no reactant in excess.
4. To calculate the grams of aluminum sulfate produced, we need to use the stoichiometry of the balanced equation.
From the balanced equation, we can see that the stoichiometric ratio between aluminum hydroxide and aluminum sulfate is 2:1. This means that 2 moles of aluminum hydroxide produce 1 mole of aluminum sulfate.
Given that the moles of aluminum hydroxide is 0.119625 mol, we can calculate the moles of aluminum sulfate produced:
Moles of aluminum sulfate = 0.119625 mol × (1 mole of aluminum sulfate / 2 moles of aluminum hydroxide)
= 0.0598125 mol
Next, we can calculate the grams of aluminum sulfate produced:
Grams of aluminum sulfate = Moles of aluminum sulfate × Molar mass of aluminum sulfate
= 0.0598125 mol × (342 g/mol)
= 20.4575 g
Therefore, approximately 20.46 grams of aluminum sulfate will be produced.