Some sulfuric acid is spilled on a lab bench. It can be neutralized by sprinklin

Question

Some sulfuric acid is spilled on a lab bench. It can be neutralized by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodium bicarbonate reacts with sulfuric acid as follows: 2 NaHCO_3(s) + H_2SO_4(aq) rightarrow Na_2SO_4(aq) + 2 H_2O(l) + 2 CO_2 (g) sodium bicarbonate is added until the fizzing due to the formation of CO_2 (g0 stops. If 27 mL of 6.0 M H_2SO_4 was spilled, what is the minimum mass of NaHCO_3 that must be added to the spill to neutralize the acid?

Explanation / Answer

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

To find the mass of sodium bicarbonate required, we need to calculate the number of moles,
which can be found from its mole ratio with sulfuric acid.

Moles of sulfuric acid = Concentration x Volume in L = 6.0 M x 0.027 L = 0.162 mol

Since the mole ratio of H2SO4 to NaHCO is 1:2 ;

Moles of sodium bicarbonate = 2 x Moles of sulfuric acid = 0.324mol
Mass of sodium bicarbonate = Molar mass x Moles = 84 x 0.324 = 27.216 g

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