Question

How many moles of sodium bicarbonate are needed to neutralise 0.8 ml of sulphuric acid at stp

Answer 1

Let's consider that 0.8 ml is 100% pure and now we will use the density of pure 100% sulphuric acid which is 1.8391 g/ml

mass is 0.8 x 1.8391 = 1.471 g

moles of H₂SO₄ = 1.471/98 = 0.015 moles.

equation for reaction 2 NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂

ratio of mole 2:1

so moles  of 2NaHCO₃ is 2 x 0.015  = 0.03 moles.

Ans:- 0.03 moles of sodium bicarbonate are needed to neutralise 0.8 ml of sulphuric acid at stp.

Answer 2

Since at STP, the concentration of any solution is alway 1m. So at STP, the concentration of H2SO4 will be 1M.

This means that 1000ml of H2SO4 solution will contain 1 mole.

Therefore 0.8ml of H2SO4 will contain 1x0.8/1000 = 0.0008moles.

The equation for the reaction is

2NaHCO3(aq) +H2SO4 (aq) ———-> Na2SO4 (aq) + H2O (l) +CO2 (g)

From the above equation we observe that 1 mole of H2SO4 reacts with 2 moles NaHCO3

Since the mole ratio is 2:1 and the moles in 0.8ml of H2SO4 are 0.0008moles, we would require exactly 2 x 0.0008moles (1.6x10–3 moles) of NaHCO3 to neutralize 0.8ml of H2SO4