Question

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 formula of molarity is:

`molarity = (number of moles of solute)/(Volume of solvent in L)` -(1)

Molarity of
`H_2SO_4 = 6.0 M` (given)

Volume of
`H_2SO_4 = 27.0 mL` (given)

Since,
`1 L = 1000 mL`

So,
`27 mL = 0.027 L`

Substituting the values in formula (1):

`6.0 M = (number of moles of H_2SO_4)/(0.027 L)`

`number of moles of H_2SO_4 = 0.027 L* 6.0 mol/L  = 0.162 mol`

The balanced chemical reaction between
`H_2SO_4` and
`NaHCO_3` is:

`H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O`

From the above reaction it is clear that, 2 moles of
`NaHCO_3` reacts with 1 mole of
`H_2SO_4`. So, for
`0.162 mol` of
`H_2SO_4` number of moles of
`NaHCO_3` is:

Number of moles of
`NaHCO_3` =
`0.162 * 2 = 0.324 mol`

Since,
`Moles = (mass)/(Molar Mass)` -(2)

Molar mass of
`NaHCO_3` =
`23 + 1+12+(3* 16) = 84 g/mol`

Using formula (2):

Mass of
`NaHCO_3` = Moles of
`NaHCO_3*` Molar mass of
`NaHCO_3`

Substituting the values:

Mass of
`NaHCO_3` =
`0.324 mol* 84g/mol = 27.216 g`

Hence,
`27.216 g` of
`NaHCO_3` must be added to neutralize the spill acid.

Answer 2

The minimum mass of NaHCO3 that must be added to the spill to neutralize the acid is 27.216 grams

calculation

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