The amount of sulphuric acid (in gm) required to neutralise 10L of ammonia at STP is
To find the amount of sulphuric acid required to neutralize ammonia, we first need to know the balanced chemical equation of the reaction.
The reaction between sulphuric acid (H₂SO₄) and ammonia (NH₃) can be represented as follows:
H₂SO₄ + 2NH₃ → (NH₄)₂SO₄
From the equation, we can see that 1 mole of sulphuric acid reacts with 2 moles of ammonia to produce 1 mole of ammonium sulfate.
To calculate the amount of sulphuric acid required, we need to follow these steps:
Step 1: Find the molar volume of ammonia at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, the molar volume of ammonia (NH₃) is 22.4 L/mol.
Step 2: Convert 10 liters of ammonia to moles
Divide the given volume of ammonia (10 L) by the molar volume of ammonia (22.4 L/mol) to get the number of moles:
10 L / 22.4 L/mol = 0.4464 mol
Step 3: Calculate the moles of sulphuric acid required
Since the stoichiometric ratio between sulphuric acid and ammonia is 1:2, we need twice as many moles of ammonia to neutralize the sulphuric acid. Therefore, the moles of sulphuric acid required would be:
0.4464 mol / 2 = 0.2232 mol
Step 4: Convert moles of sulphuric acid to grams
To convert moles to grams, we need to know the molar mass of sulphuric acid, which is 98.09 g/mol. Multiply the moles of sulphuric acid by its molar mass to get the mass in grams:
0.2232 mol * 98.09 g/mol = 21.96 g
Therefore, approximately 21.96 grams of sulphuric acid is required to neutralize 10 liters of ammonia at STP.