Chapter 3, Problem 3.132QP

Interpretation Introduction

Interpretation:

The required mass of Ammonium nitrate to produce $\text{685}\text{mg}$ of ${\text{N}}_{\text{2}}\text{O}$ in $\text{75\%}$ of the reaction should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

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