Question

# Determine how many milliliters of 0.216 M HClO4 will be required to neutralize 36.61 g Ca(OH)2...

Determine how many milliliters of 0.216 M HClO4 will be required to neutralize 36.61 g Ca(OH)2 according to the reaction: _____ × 10 ______mL Determine how many grams of Al(OH)3 will be required to neutralize 237 mL of 0.400 M HCl according to the reaction: 3HCl(aq) + Al(OH)3(s) right-arrowAlCl3(aq) + 3H2O(l) __________________g

2HClO4(aq) + Ca(OH)2(aq) ----> Ca(ClO4)2 + 2H2O(l)

molar mass of Ca(OH)2 = 74 g/mole

moles of Ca(OH)2 in 36.61 g of it = mass/molar mass = 36.61/74 = 0.495

as per the balanced reaction; HClO4 & Ca(OH)2 = 2:1

Thus, moles of HClO4 required = 2*moles of Ca(OH)2 reacting = 2*0.495 = 0.990

Hence, volume of HClO4 required = moles/molar concentration = 0.99/0.216 = 4.581 litres = 4.581*103 ml

2) 3HCl(aq) + Al(OH)3(s) -------> AlCl3(aq) + 3H2O(l)

moles of HCl reacting = molarity*volume of solution in litres = 0.4*0.237 = 0.0948

As per the balanced reaction HCl & Al(OH)3 reacts in the molar ratio of 3:1

Thus, moles of Al(OH)3 reacting = (1/3)*moles of HCl reacting = 0.0316

Molar mass of Al(OH)3 = 78 g/mole

Thus, mass of Al(OH)3 reacting = moles*molar mass = 2.4648g

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