# Answer to Question #204127 in General Chemistry for Jasmine Enriquez

Question #204127

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?

1
2021-06-07T03:13:25-0400

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

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