Answer to Question #96121 in General Chemistry for john

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 resulting solution. The sodium bicarbonate reacts with sulfuric acid according to:

2 NaHCO3(s)+H2SO4(aq)−→  Na2SO4(aq)+2 H2O(l)+2 CO2(g)
Sodium bicarbonate is added until the fizzing due to the formation of CO2(g) stops. If 27 mL of 6.0 M H2SO4 was spilled, what is the minimum mass of NaHCO3 that must be added to the spill to neutralize the acid?

Answer 1

The amount in moles of spilled sulfuric acid is

$n(H_2SO_4) = c(H_2SO_4) \cdot V(H_2SO_4)$

According to stoichiometry of the chemical reaction

$2NaHCO_{3(s)}+H_2SO_{4(aq)} \rightarrow Na_2SO_{4(aq)}+2H_2O_{(l)}+2CO_{2(g)}$

the amount of sodium bicarbonate is two times of the amount of sulfuric acid. Thus

$n(NaHCO_3) = 2 \cdot n(H_2SO_4)$

The mass of sodium bicarbonate can be calculated:

$m(NaHCO_3) = n(NaHCO_3) \cdot M(NaHCO_3) = 2 \cdot c(H_2SO_4) \cdot V(H_2SO_4) \cdot M(NaHCO_3) = 2 \cdot 6.0 \frac{mol}{L} \cdot 0.027 L \cdot 84 \frac{g}{mol} = 27g$