Chapter 4, Problem 4.71QP

Interpretation Introduction

Interpretation:

The mass of $\text{AgCl}$ precipitate formed from the reaction of ${\text{AgNO}}_{\text{3}}$ and ${\text{CaCl}}_{\text{2}}$ has to be calculated.

Concept introduction:

Precipitation reaction:

When the two solutions mixed together and an insoluble compound settles down in the solution as a precipitate is called precipitation reaction.

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

Mole:

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

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

$\text{Molarity = }\frac{\text{No}\text{. of moles of solute}}{\text{Volume of solution in L}}$

Mass:

The mass of solute present in the solution is given by multiplication of mole and molar mass of the solute.

$\text{Mass}\text{=}\text{Mole}\text{×}\text{Molar}\text{mass}$

Limiting reagent:

The reagent, which is completely consumed in the hole reaction is called limiting reagent and the amount of this reagent is limited to the amount of produced formation.