Question

If 35 ml of 6.0 m h2so4 was spilled, calculate the minimum mass of nahco3 that must be added to the spill to neutralize the acid.

Answer 1

The balanced equation between the
`H_2SO_4` and
`NaHCO_3` is:

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

Formula of molarity is:

`Molarity = (Moles of solute)/(Volume of solution in Liters)`

`Molarity = 6.0 M, Volume = 35 mL = 0.035 L`

Substituting the values,

`6 = (Moles of solute)/(0.035)`

`Moles of solute = 0.035 L* 6 mol/L = 0.21 mole`

So, number of moles of
`H_2SO_4` is
`0.21 mole`.

From the balanced equation it is clear that for 1 mole of
`H_2SO_4`, 2 moles of
`NaHCO_3` are required.

Hence, 0.21 mole of
`H_2SO_4` =
`2* 0.21 mole = 0.42 mole of NaHCO_3`

Molar mass of
`NaHCO_3` =
`84.007 g/mol`

So, the mass of
`NaHCO_3` =
`84.007 g/mol * 0.42 mol = 35.283 g`