Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by spr

Question

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodium bicarbonate reacts with sulfuric acid according to: 2 NaHCO_3(s) + H_2SO_4(aq) rightarrow Na_2SO_4(aq) + 2H_2O(l) + 2 CO_2(g) Sodium bicarbonate is added until the fizzing due to the formation of CO_2(g) 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(s) + H2SO4 ------> Na2SO4(aq) + 2H2O(l) + 2CO2

According to stoichiometric coefficient of equation 2 moles of NaHCO3 completely neutralize1 mole of H2SO4.

Molarity = Moles of solute / Volume of solution in litres

Molarity*Volume of solution in litres = Moles of solute

Molarity of H2SO4 = 6.0M

Volume of H2SO4 solution = 27mL or 0.027L

6.0*0.027 = 0.162 = Moles of H2SO4

So number of moles of NaHCO3 required to neutralize 0.162 moles of H2SO4 = 2*0.162 =0.324

Moles = Given mass / Molar mass

Molar mass of NaHCO3 = 23 + 1 + 12 +3*16 =84 g/mol

0.324 = Given mass/ 84

Given mass = 27.216 g

27.216 grams of NaHCO3 is required to completely neutralize 27mL of 6.0M H2SO4.

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