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

Jun 8, 2016

$20.4 g$ $N a O H$

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:

${H}_{2} S {O}_{4} + 2 N a O H \rightarrow N {a}_{2} S {O}_{4} + 2 {H}_{2} O$

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 $98.08 \frac{g}{m o l}$ represents the molar mass of sulfuric acid and the $40.0 \frac{g}{m o l}$ is the molar mass of sodium hydroxide.

So, what we have is:

$g N a O H$ = $25.0 g {H}_{2} S {O}_{4} \times \frac{1 m o l {H}_{2} S {O}_{4}}{98.08 g {H}_{2} S {O}_{4}} \times \frac{2 m o l N a O H}{1 m o l {H}_{2} S {O}_{4}} \times \frac{40.0 g N a O H}{1 m o l N a O H}$ $=$ $20.4 g$ $N a O H$