what volume of NH3 gas at s.t.p. is required to neutralise 40cm3 of 2M H2SO4?
To find the volume of NH3 gas required to neutralize 40 cm3 of 2M H2SO4 at standard temperature and pressure (STP), we need to use the balanced chemical equation for the reaction between NH3 and H2SO4.
The balanced chemical equation for the reaction is:
2 NH3 + H2SO4 -> (NH4)2SO4
From the balanced equation, we can see that the stoichiometric ratio between NH3 and H2SO4 is 2:1. This means that for every 2 moles of NH3, 1 mole of H2SO4 is required for complete neutralization.
Step 1: Convert the given volume of H2SO4 to moles.
40 cm3 of 2M H2SO4 means 2 moles of H2SO4 per 1 liter of solution.
So, 40 cm3 is equal to 0.04 liters.
Moles of H2SO4 = Volume (liters) x Concentration (Molarity)
Moles of H2SO4 = 0.04 liters x 2 M = 0.08 moles
Step 2: Determine the moles of NH3 required.
Since the stoichiometric ratio between NH3 and H2SO4 is 2:1, the moles of NH3 required will be half of the moles of H2SO4.
Moles of NH3 = Moles of H2SO4 / 2
Moles of NH3 = 0.08 moles / 2 = 0.04 moles
Step 3: Convert moles of NH3 to volume at STP.
1 mole of any gas occupies 22.4 liters at STP. Therefore, 0.04 moles of NH3 will occupy:
Volume of NH3 = Moles of NH3 x 22.4 liters/mol
Volume of NH3 = 0.04 moles x 22.4 liters/mol = 0.896 liters
Therefore, the volume of NH3 gas required to neutralize 40 cm3 of 2M H2SO4 at STP is approximately 0.896 liters.