The minimum mass of NaHCO3 that must be added to the spill to neutralize the acid is 27.216 grams
write the balanced chemical equation
2NaHCO3 +H2SO4 → Na2SO4 +2H2O +2CO2
find the moles of H2SO4 = molarity x volume in liters
volume in liters = 27/1000=0.027 l
moles is therefore= 0.027 x6=0.162 moles
by use of mole ratio of NaHCO3: H2SO4 which is 2:1 the moles of NaHCO3=0.162 x2=0.324 moles
mass of NaHCO3= moles of NaHCO3 x molar mass of NaHCO3(84g/mol)
= 84g/mol x 0.324=27.216 grams
The formula of molarity is:
[tex]molarity = \frac{number of moles of solute}{Volume of solvent in L}[/tex] -(1)
Molarity of [tex]H_2SO_4 = 6.0 M[/tex] (given)
Volume of [tex]H_2SO_4 = 27.0 mL[/tex] (given)
Since, [tex]1 L = 1000 mL[/tex]
So, [tex]27 mL = 0.027 L[/tex]
Substituting the values in formula (1):
[tex]6.0 M = \frac{number of moles of H_2SO_4}{0.027 L}[/tex]
[tex]number of moles of H_2SO_4 = 0.027 L\times 6.0 mol/L = 0.162 mol[/tex]
The balanced chemical reaction between [tex]H_2SO_4[/tex] and [tex]NaHCO_3[/tex] is:
[tex]H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O[/tex]
From the above reaction it is clear that, 2 moles of [tex]NaHCO_3[/tex] reacts with 1 mole of [tex]H_2SO_4[/tex]. So, for [tex]0.162 mol[/tex] of [tex]H_2SO_4[/tex] number of moles of [tex]NaHCO_3[/tex] is:
Number of moles of [tex]NaHCO_3[/tex] = [tex]0.162 \times 2 = 0.324 mol[/tex]
Since, [tex]Moles = \frac{mass}{Molar Mass}[/tex] -(2)
Molar mass of [tex]NaHCO_3[/tex] = [tex]23 + 1+12+(3\times 16) = 84 g/mol[/tex]
Using formula (2):
Mass of [tex]NaHCO_3[/tex] = Moles of [tex]NaHCO_3\times[/tex] Molar mass of [tex]NaHCO_3[/tex]
Substituting the values:
Mass of [tex]NaHCO_3[/tex] = [tex]0.324 mol\times 84g/mol = 27.216 g[/tex]
Hence, [tex]27.216 g[/tex] of [tex]NaHCO_3[/tex] must be added to neutralize the spill acid.
