Question

Sodium Bicarbonate is sometimes used to neutralize acid spills. If 2.0L of 6.0M H2SO4 is spilled, how many liters of Sodium Bicarbnate, NaHCO3, is required to neutralize H2SO4?

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:

$m(NaHCO_3) = n(NaHCO_3) \cdot M(NaHCO_3) = 2 \cdot c(H_2SO_4) \cdot V(H_2SO_4) \cdot M(NaHCO_3) = 2 \cdot 6.0 \frac{mol}{L} \cdot 2 L \cdot 84 \frac{g}{mol} = 2016g$

no. of moles of NaHCO₃ = mass/molar mass

= 2016/84.007 = 23.99 moles

Volume required :-

Molarity = no. of moles of solute/ volume (L)

6.0M = 23.99/volume(L)

Volume(L) = 23.99/6 = 3.99 L