If 27mL of 6.0 M sulfuric acid was spilled, what is the minimum mass of sodium bicarbonate that must be added to the spill to neutralize the acid?

Answer:- 27.2 g

Solution:- The balanced equation for the reaction of sulfuric acid with sodium bicarbonate is:

[tex]H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O[/tex]

From this equation, one mole of sulfuric acid reacts with two moles of sodium bicarbonate. We could calculate the moles of sulfuric acid from it's given volume and molarity. Then using mol ratio, we could calculate the moles of sodium bicarbonate required. Moles could easily be converted to grams on multiplying by molar mass.

The calculations would be shown as:

[tex]27mL(\frac{1L}{1000mL})(\frac{6.0moleH_2SO_4}{1L})(\frac{2moleNaHCO_3}{1moleH_2SO_4})(\frac{84gNaHCO_3}{1moleNaHCO_3})[/tex]

= [tex]27.2gNaHCO_3[/tex]

So, 27.2 g of sodium bicarbonate are required to neutralize the given sulfuric acid solution.