Chapter 19, Problem 120A

Interpretation Introduction

Interpretation : The grams of oxygen are to be calculated.

Concept Introduction : Calculating how much reactant is required or how much product will be produced in a reaction is based on the balanced chemical equation.

The formula to calculate the number of moles is given as

  $\text{no}\text{.}\text{of}\text{moles}\left(n\right)=\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Answer to Problem 120A

The grams of oxygen is $130.9\text{g}$ .

Explanation of Solution

The given mass of sulfur is $87.4\text{g}$ .

Stoichiometry is a branch of chemistry that deals with calculating the quantities involved in chemical reactions.

The number of atoms, molecules, or moles; mass; and volume are some examples of different parameters that can be used to interpret a balanced chemical equation.

The given equation to produce sulfur trioxide in balanced form is given as

  $\text{2S}\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{SO}}_{\text{3}}\left(g\right)$

The stoichiometry of sulfur and oxygen is in $3:2$ .

The number of moles is given as

  $\text{no}\text{.}\text{of}\text{moles}\left(n\right)=\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

The molecular mass of sulfur is $32.06\text{g/mol}$ .

To calculate the number of moles of sulfur, substitute the values in the above formula:

  $\begin{array}{c}\text{no}\text{.}\text{of}\text{moles}\left(n\right)=\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}\\ =\frac{87.4\text{g}}{32.06\text{g/mol}}\\ =2.73\text{mol}\end{array}$

The stoichiometry between sulfur and oxygen can be written as

  $\begin{array}{l}\frac{n\left({\text{O}}_{\text{2}}\right)}{n\left(\text{S}\right)}=\frac{3}{2}\\ n\left({\text{O}}_{\text{2}}\right)=\frac{3}{2}\times n\left(\text{S}\right)\end{array}$

To calculate the number of moles of oxygen, substitute the values in above formula:

  $\begin{array}{c}n\left({\text{O}}_{\text{2}}\right)=\frac{3}{2}\times n\left(\text{S}\right)\\ =\frac{3}{2}\times 2.73\text{mol}\\ \text{=}\text{4}\text{.09}\text{mol}\end{array}$

To calculate the mass of oxygen, substitute the values in number of moles formula:

  $\begin{array}{c}\text{no}\text{.}\text{of}\text{moles}\left(n\right)=\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}\\ 4.09\text{mol}=\frac{\text{mass}\text{of}{\text{O}}_2}{32\text{g/mol}}\\ \text{mass}\text{of}{\text{O}}_{\text{2}}=4.09\text{mol}\times 32\text{g/mol}\\ =130.9\text{g}\end{array}$