Some sulfuric acid (H2SO4) is spilled on a lab bench. It can be neutralized by sprinkling sodium bicarbonate (NaHCO3) on it creating liquid water (H2O), carbon dioxide (CO2) gas and aqueous sodium...

Some sulfuric acid (H2SO4) is spilled on a lab bench. It can be neutralized by sprinkling sodium bicarbonate (NaHCO3) on it creating liquid water (H2O), carbon dioxide (CO2) gas and aqueous sodium sulfate (Na2SO4). 

If 25 mL of 6.0 M H2SO4 was spilled, how many moles of NaHCO3 must be added to neutralize the
spill?

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The balanced chemical equation for the given reaction is :

`H_2SO_4+2NaHCO_3 rarr Na_2SO_4+2CO_2+2H_2O`

`H_2 SO_4` is a di-basic acid. So, its Normality will be 2 * Molarity.

Hence, 25 ml, 6.0M `H_2SO_4` = 25ml, 12N `H_2SO_4`

=`25*12` meq. acid

=`300` meq. acid

`NaHCO_3` is a monoacidic base.

Since, acid-base neutralization reaction occurs on the basis of their equivalence a total of `300` meq. i.e `0.3` equivalence = `0.3 ` moles base will be required.

Therefore, 0.3 moles of `NaHCO_3` must be added to neutralize the spill.

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