Chapter U4.78, Problem 6E

Interpretation Introduction

Interpretation : The compounds are to be stated in the increasing order of grams.

Concept Introduction : The formula to calculate the number of moles is given as shown below.

  $\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$ (1)

Answer to Problem 6E

The compounds in increasing order of grams are as follows.

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}<{\text{SiCl}}_4<\text{PbO}$

Explanation of Solution

Given information : The given moles of ${\text{SiCl}}_4$ , $\text{PbO}$ , and ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ are $\text{2}\text{.0}\text{mol}$ .

The molar mass of ${\text{SiCl}}_4$ is $\text{169}\text{.9}\text{g}/\text{mol}$ . Substitute the given moles as $\text{2}\text{.0}\text{mol}$ and molar mass as $\text{169}\text{.9}\text{g}/\text{mol}$ for ${\text{SiCl}}_4$ in equation (1) as shown below.

  $\begin{array}{c}\text{2}\text{.0}\text{mol}=\frac{\text{Mass}}{\text{169}\text{.9}\frac{\text{g}}{\text{mol}}}\\ \text{Mass}=339.8\text{g}\end{array}$

The molar mass of $\text{PbO}$ is $\text{223}\text{.2}\text{g}/\text{mol}$ . Substitute the given moles as $\text{2}\text{.0}\text{mol}$ and molar mass as $\text{223}\text{.2}\text{g}/\text{mol}$ for $\text{PbO}$ in equation (1) as shown below.

  $\begin{array}{c}\text{2}\text{.0}\text{mol}=\frac{\text{Mass}}{\text{223}\text{.2}\frac{\text{g}}{\text{mol}}}\\ \text{Mass}=446.4\text{g}\end{array}$

The molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\text{159}\text{.69}\text{g}/\text{mol}$ . Substitute the given moles as $\text{2}\text{.0}\text{mol}$ and molar mass as $\text{159}\text{.69}\text{g}/\text{mol}$ for ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ in equation (1) as shown below.

  $\begin{array}{c}\text{2}\text{.0}\text{mol}=\frac{\text{Mass}}{159.69\frac{\text{g}}{\text{mol}}}\\ \text{Mass}=319.38\text{g}\end{array}$

Therefore, the compounds in increasing order of grams are as follows.

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}<{\text{SiCl}}_4<\text{PbO}$

Conclusion

The compounds in increasing order of grams are as follows.

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}<{\text{SiCl}}_4<\text{PbO}$