"What amount of Barium hydroxide is needed to neutralize 500 kg of 18 molar Sulphuric acid?"
To determine the amount of Barium hydroxide needed to neutralize 500 kg of 18 molar Sulphuric acid, you will need to follow these steps:
Step 1: Determine the number of moles of Sulphuric acid present in 500 kg. To do this, we need to use the formula:
Moles = Mass / Molar mass
The molar mass of Sulphuric acid (H2SO4) is:
2(1.01 g/mol for hydrogen) + 32.07 g/mol (for sulfur) + 4(16.00 g/mol for oxygen) = 98.09 g/mol
Converting the mass from kilograms to grams:
Mass = 500 kg * 1000 g/kg = 500,000 g
Calculating the number of moles:
Moles = 500,000 g / 98.09 g/mol = 5096.4 mol
Step 2: Determine the stoichiometric ratio between Sulphuric acid (H2SO4) and Barium hydroxide (Ba(OH)2) by examining their balanced chemical equation:
H2SO4 + 2Ba(OH)2 -> BaSO4 + 2H2O
From the equation, we can see that 1 mole of Sulphuric acid reacts with 2 moles of Barium hydroxide.
Step 3: Calculate the amount of Barium hydroxide needed. Since the stoichiometric ratio is 1:2 (Sulphuric acid:Barium hydroxide), you need double the number of moles of Sulphuric acid.
Number of moles of Barium hydroxide = 2 * Moles of Sulphuric acid = 2 * 5096.4 mol = 10192.8 mol
Step 4: Convert the number of moles to the amount of Barium hydroxide in grams. First, determine the molar mass of Barium hydroxide (Ba(OH)2):
(1.01 g/mol for hydrogen) + 16.00 g/mol (for oxygen) + 137.33 g/mol (for barium) = 171.33 g/mol
Calculating the mass:
Mass = Moles * Molar mass = 10192.8 mol * 171.33 g/mol = 1748.5 kg
Therefore, approximately 1748.5 kg of Barium hydroxide is needed to neutralize 500 kg of 18 molar Sulphuric acid.