How many moles of sodium bicarbonate are needed to neutralize 0.8ml of sulphuric acid at Stp

dear student,
The neutralization reaction is   2NaHCO3 + H2SO4 → Na2SO4 + 2H2CO3
as you have not provided the density of the sulphuric acid so I am assuming it as 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
regards
 

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