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)

Question

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?

Answer 1

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.