Question

How many grams of sodium hydroxide are needed to completely neutralize 25.0 grams of sulfuric acid?

Answer 1

`20.4g` `NaOH`

Explanation:

A Neutralization reaction occurs when a strong acid reacts with a strong base to produce water and a salt (ionic compound).
In our case, sulfuric acid (strong acid) will react with sodium hydroxide (strong base) to form water and sodium sulfate:

`H_2SO_4 + 2 NaOH rarr Na_2SO_4 + 2H_2O`

We start with the units that we want to end up with and we set that equal to our given value, which will be multiplied by a conversion factor (the molar ratio from the balanced chemical equation). The `98.08g/(mol)` represents the molar mass of sulfuric acid and the `40.0 g/(mol)` is the molar mass of sodium hydroxide.

So, what we have is:

`g NaOH` = `25.0 gH_2SO_4xx (1 mol H_2SO_4)/(98.08g H_2SO_4)xx(2mol NaOH)/(1 mol H_2SO_4)xx(40.0g NaOH)/(1mol NaOH)` `=` `20.4g` `NaOH`