Question

How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?

Answer 1

Two moles.

Step-by-step explanation:

Sulphuric (sulfuric) acid
`\rm H_2SO_4` is a diprotic acid. When one mole of
`\rm H_2SO_4` molecules dissolve in water, two moles of
`\rm H^(+)` ions would be produced.

`\rm H_2SO_4 \to 2\, H^(+) + {SO_4}^(2-)`.

On the other hand, sodium hydroxide
`\rm NaOH` is a monoprotic base. When one mole of
`\rm NaOH` formula units dissolve in water, only one mole of hydroxide ions
`\rm OH^(-)` would be produced.

`\rm NaOH \to Na^(+) + OH^(-)`.

Note that
`\rm H^(+)` and
`\rm OH^(-)` react at a one-to-one ratio:

`\rm H^(+) + OH^(-) \to H_2O`.

As a result, it would take
`2\; \rm mol` of
`\rm OH^(-)` to react with the
`\rm 2\; mol` of
`\rm H^(+)` that was released when
`1\; \rm mol` of
`\rm H_2SO_4` is dissolved in water. Since one mole of
`\rm NaOH` formula units could produce only one mole of
`\rm OH^(-)`, it would take
`\rm 2\; mol` of
`\rm NaOH` formula units to produce that
`2\; \rm mol` of
`\rm OH^(-)` for reacting with
`1\; \rm mol` of
`\rm H_2SO_4`.