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Explain Bot 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.