Question

how many mole of sodium bicarbonate are needed to neutrilize 0.08 ml of sulphuric acid​

Answer 1

Answer:

Explanation:

The density of 100% concentrated sulfuric acid can be looked up and is 1.8391 g/mL. So, with 0.8 mL you would have:

0.8 mL * 1.8391 g/mL = 1.471 g

Now, convert this to moles

1.471 g / 98 g/mol H2SO4 = 0.0150 moles

The neutralization reaction is

2NaHCO3 + H2SO4 → Na2SO4 + 2H2CO3

2H2CO3 → 2H2O + 2CO2

Overall, the reaction is

2NaHCO3 + H2SO4 → Na2SO4 + 2H2O + 2CO2

The stoichiometry says you need two moles of NaHCO3 for every mole of sulfuric acid. So, you will need

0.0150 mole H2SO4 * 2 NaHCO3/1H2SO4 = 0.03 mole NaHCO3

Answer 2

The answer is 0.0028 moles.

GIVEN

Volume of sulphuric acid = 0.08 ml

TO FIND

Moles of sodium carbonate to neutralise 0.08ml of sulphuric acid.

SOLUTION

We can simply solve the above problem as follows;

Volume of sulphuric acid = 0.08ml

Density of sulphuric acid = 1.83 g/cm³

we know that,

Density = Mass/Volume

Mass = Density × Volume = 1.83 × 0.08 = 0.14 grams

Moles of H₂SO₄ = 0.14/98 = 0.0014 moles.

Now,

Neutralisation reaction of H₂SO₄

H₂SO₄ + 2NaHCO₃ --------> Na₂SO₄ + H₂O + CO₂

We can observe that 2 moles of NaHCO₃ are required for 1 mole of H₂SO₄.

Therefore,

0.0014 moles of H₂SO₄ will require, 2 × 0.0014 = 0.0028 moles.

Hence, The answer is 0.0028 moles.

#Sp1