Answer:
What is the maximum amount of calcium sulfate that can be formed? 53.1 grams CaSO4
What is the FORMULA for the limiting reagent? H2SO4
What amount of the excess reagent remains after the reaction is complete? 4.59 grams of Ca(OH)2
Explanation:
Step 1: Data given
Mass of sulfuric acid = 38.3 grams
Molar mass of H2SO4 = 98.08 g/mol
Mass of calcium hydroxide = 33.5 grams
Molar mass of Ca(OH)2 = 74.09 g/mol
Step 2: The balanced equation
H2SO4 + Ca(OH)2 → CaSO4 + 2H2O
Step 3: Calculate moles of H2SO4
moles H2SO4 = mass H2SO4 / molar mass H2SO4
moles H2SO4 = 38.3 grams / 98.08 g/mol
moles H2SO4 = 0.390 moles
Step 4: Calculate moles of Ca(OH)2
moles Ca(OH)2 = 33.5 grams / 74.09 g/mol
moles Ca(OH)2 =0.452 moles
Step 5: Calculate limiting reactant
For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O
H2SO4 is the limiting reactant. It will completely be consumed (0.390 moles).
Ca(OH)2 is in excess. There will be consumed 0.390 moles
There will remain 0.452 - 0.390 = 0.062 moles
This is 0.062 * 74.09 g/mol = 4.59 grams
Step 6: Calculate moles of calcium sulfate
For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O
For 0.390 moles of H2SO4, there will be produced 0.390 moles of CaSO4
Step 7: Calculate mass of CaSO4
Mass CaSO4 = moles CaSO4 * molar mass CaSO4
Mass CaSO4 = 0.390 moles * 136.14 g/mol
Mass of CaSO4 = 53.1 grams
