Chapter 9, Problem 9.123QP

Interpretation Introduction

Interpretation:

The produced mass of ${\text{BaSO}}_{\text{4}}$ precipitate, when reaction between given ${\text{Ba(OH)}}_{\text{2}}$ and ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ solutions should be calculated

Concept introduction:

Precipitation reaction:

If a precipitate is formed, when the two solutions mixed together is called precipitation reaction.

The mass of the precipitate is depends on the reactant masses.

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

Mass:

The mass of compound is given by the multiplication of mole of the compound and molar mass of the compound.

$\text{Mass }\text{=}\text{Mole}\text{×}\text{Molar}\text{mass}\text{(g)}$