Chapter 3, Problem 3.77P

Interpretation Introduction

Interpretation:Moles of excess reactant for reaction of $\text{350}\text{.0 mL}$ of $\text{0}{\text{.210 M H}}_2{\text{SO}}_{\text{4}}$ and $\text{0}\text{.500 L}$ of $\text{0}\text{.196 M NaOH}$ is to be calculated.

Concept introduction: The formula to evaluate moles from molarity is given as follows:

  $n=MV$

Where,

• $V$represents volume.
• $M$represents molarity.
• $n$represents number of moles.

The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{molar mass}\right)$