Answer:
The base will neutralize all of the acid. There will remain 12.0 grams of Mg(OH)2 in excess.
Step-by-step explanation:
Step 1: Data given
Mass of Hydrochloric acid (HCl) = 20.0 grams
Mass of Magnesium hydroxide (Mg(OH)2)
Molar mass of HCl = 36.46 g/mol
Molar mass of Mg(OH)2 = 58.32 g/mol
Step 2: The balanced equation
2HCL + Mg(OH)2 → MgCl2 + 2H2O
Step 3: Calculate moles of HCl
Moles HCl = 20.00 grams / 36.46 g/mol
Moles HCl = 0.5485
Step 4: Calculate moles of Mg(OH)2
Moles Mg(OH)2 = 28.00 grams / 58.32 g/mol
Moles Mg(OH)2 = 0.4801 moles
Step 5: Determine the limiting reactant
For 2 moles HCl we need 1 mol of Mg(OH)2 to produce 1 mol of MgCl2 and 2 moles of H2O
HCl is the limiting reactant. It will completely be consumed (0.5485 moles).
Mg(OH)2 is in excess. There will react 0.5485/2 = 0.27425 moles
There will remain 0.4801 - 0.27425 = 0.20585 moles of Mg(OH)2
Remaining mass of Mg(OH)2 = 0.20585 moles * 58.32 g/mol = 12.0 grams
The base will neutralize all of the acid. There will remain 12.0 grams of Mg(OH)2 in excess.