Answer to Question #204127 in General Chemistry for Jasmine Enriquez

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