Chapter 4.9, Problem 4.9Q

(a)

Interpretation Introduction

Interpretation:

The amount (in $\text{g}$) of ${\text{NO}}_{\text{2}}$ that is produced from depletion of $\text{50}{\text{.0 g O}}_3$ has to be calculated.

Concept Introduction:

Moles:

Mole of the compound is found by dividing mass of the substance with its molar mass.

  $\text{No}\text{. of moles (n) = }\frac{\text{mass}}{\text{Molar mass}}$

Mass:

Mass of the compound is calculated by mole of the compound multiplied with molar mass of the compound.

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

(b)

Interpretation Introduction

Interpretation:

The $\%$ yield of ${\text{NO}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Percent yield:

The division of actual yield by the theoretical yield multiplied by $100\%$ gives the percent yield of a reaction.

  $\text{Percent yield = }\frac{\text{Actual yield of product}}{\text{Theoretical yield of product}}\times 100\%$