Question

If 28 ml of 5.4 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?

Answer 1

A reaction between acid and base to form water and salt is known as neutralization reaction. It is a double replacement reaction.

The reaction between
`H_2SO_4` and
`NaHCO_3` will be:

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

The balanced reaction is:

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

Volume of
`H_2SO_4` = 28 mL (given)

Since, 1 L = 1000 mL

So, 28 mL = 0.028 L

Molarity of
`H_2SO_4` = 5.4 M (given)

Molarity =
`(number of moles of solute)/(Volume of solution in Liters)` -(1)

Substituting the values in equation (1):

`5.4 mol/L = (number of moles of H_2SO_4)/(0.028 L)`

`number of moles of H_2SO_4 = 5.4 mol/L*0.028 L`

`number of moles of H_2SO_4 = 0.1512 mol`

From the balanced reaction between
`H_2SO_4` and
`NaHCO_3`, 2 moles of
`NaHCO_3` reacts with 1 mole of
`H_2SO_4`.

Molar mass of
`NaHCO_3` = 84.007 g/mol

Mass of
`NaHCO_3` needed:

`0.1512 mol H_2SO_4 * (2 mole NaHCO_3)/(1 mole H_2SO_4)* (84.007 g)/(1 mole NaHCO_3)` =
`25.404 g`

Hence, the required amount of
`NaHCO_3` is
`25.404 g`.