How many mols of #H_2SO_4# will neutralize 78.3 mK of 0.187 N #NaOH#?

Question

Natalie Anton · Accepted Answer

See below.

For this calculation, I'm assuming you meant 78.3 mL (not "mK") and 0.187M NaOH (basically that you made two major typos, since both K and N are close together to letters l and M which would make sense on a standardized keyboard).

If that is the case, and the units are not arbitrary, then the calculation should be correct. If it's not then I'm sorry as this is the best I could do with the information provided.

If this is the type of question you are trying to answer, remember to always start with a balanced chemical equation. The chemical system in question must be balanced to perform the calculation.

In our case:

#H_2SO_4 + 2NaOH -> 2H_2O + Na_2(SO_4)#

and

Consider the graphic below:

Jeremiah Adams · Answer

undefined To find the number of moles of $ \text{H}_2\text{SO}_4 $ required to neutralize $ 78.3 \, \text{mL} $ of $ 0.187 \, \text{N} $ $ \text{NaOH} $, you can use the equation:

$$
\text{moles of } \text{H}_2\text{SO}_4 = \text{volume of } \text{NaOH} \times \text{normality of } \text{NaOH}
$$

Substitute the given values:

$$
\text{moles of } \text{H}_2\text{SO}_4 = 78.3 \times 0.187
$$

$$
\text{moles of } \text{H}_2\text{SO}_4 = 14.6291
$$

So, approximately $ 14.63 \, \text{moles} $ of $ \text{H}_2\text{SO}_4 $ will neutralize $ 78.3 \, \text{mL} $ of $ 0.187 \, \text{N} $ $ \text{NaOH} $.