Chapter 11, Problem 11.85QP

Interpretation Introduction

Interpretation:

The given reaction data equation, formulas for the oxide and Sulfate of the metal M has to be determined.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T.  An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$\text{PV = nRT}$

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge are the similar for both the reactants and the products.

Answer to Problem 11.85QP

The mole ratio of ${\text{H}}_{\text{2}}\text{ to M is 1}\text{.5:1}\text{.}$

${\text{M}}_{\left(\text{s}\right)}{\text{+ 3HCl}}_{\left(\text{aq}\right)}\to \text{1}{\text{.5H}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{ + MCl}}_{\text{3}\left(\text{aq}\right)}$

Therefore, the formulas of the metal oxide and the Metal Sulfate are ${\text{M}}_{\text{2}}\text{O}\text{and}{\text{M}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$, respectively.

Explanation of Solution

To calculate the moles of the metal consumed in the reaction

$\text{mol}\text{M}\text{=}\text{0}\text{.225}\text{g}\text{M}\text{×}\frac{\text{1}\text{mol}\text{M}}{\text{27}\text{.0}\text{g}\text{M}}\text{=}\text{8}\text{.33}\text{×}{\text{10}}^{\text{-3}}\text{mol}\text{M}$

The moles of metal consume is calculated by plugging in the values of the given weight of metal and molecular weight of metal.  The moles of metal consume was found to be $\text{8}\text{.33}\text{×}{\text{10}}^{\text{-3}}\text{mol}\text{M}$

To calculate the moles of ${\text{H}}_{\text{2}}$ gas using the ideal gas equation.

$\begin{array}{l}\text{P}\text{=}\text{(741}\text{mmHg)}\left(\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\right)\text{=}\text{0}\text{.975}\text{atm}\\ \text{T}\text{=}\text{17°C}\text{+}\text{273}\text{=290K}\end{array}$

$\begin{array}{l}{\text{ n}}_{{\text{H}}_{\text{2}}}\text{=}\frac{{\text{P}}_{{\text{H}}_{\text{2}}}\text{V}}{\text{RT}}\\ {\text{ n}}_{{\text{H}}_{\text{2}}}\text{=}\frac{\text{(0}\text{.975}\text{atm}\text{)(0}\text{.303}\text{L)}}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(290K)}}\text{=}\text{1}\text{.24}\text{×}{\text{10}}^{\text{-2}}\text{mol}{\text{H}}_{\text{2}}\end{array}$

The moles of ${\text{H}}_{\text{2}}$ gas is calculated by plugging in the values of the given volume of ${\text{H}}_{\text{2}}$ gas, temperature and pressure.  The moles of ${\text{H}}_{\text{2}}$ gas was found to be $\text{0}\text{.303}\text{mol}{\text{H}}_{\text{2}}$

Compare the number of moles of ${\text{H}}_{\text{2}}$ produced to the number of moles of M consumed.

$\frac{\text{1}\text{.24}\text{×}{\text{10}}^{\text{-2}}\text{mol}{\text{H}}_{\text{2}}}{\text{8}\text{.33}\text{×}{\text{10}}^{\text{-3}}\text{mol}\text{M}}\approx \text{1}\text{.5}$

This means that the mole ratio of ${\text{H}}_{\text{2}}\text{ to M is 1}\text{.5:1}\text{.}$

We can now write the balanced equation since we know the mole ratio between ${\text{H}}_{\text{2}}\text{and}\text{M}$

The unbalanced equation is

$\text{M}\left(\text{s}\right)\text{ + HCl}\left(\text{aq}\right)\to \text{1}{\text{.5H}}_{\text{2}\left(\text{g}\right)}{\text{ + M}}_{\text{x}}{\text{Cl}}_{\text{y}}{}_{\left(\text{aq}\right)}$

We have three atoms of H on the products side of the reaction, so a 3 must be placed in front of $\text{HCl}$. The ratio of $\text{M to Cl}$ on the reactants side is now 1:3. Therefore the formula of the metal chloride must be $\text{MCl3}\text{.}$

The balanced equation is

${\text{M}}_{\left(\text{s}\right)}{\text{+ 3HCl}}_{\left(\text{aq}\right)}\to \text{1}{\text{.5H}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{ + MCl}}_{\text{3}\left(\text{aq}\right)}$

From the formula of the metal chloride, we determine that the charge of the metal is

+3. Therefore, the formulas of the metal oxide and the Metal Sulfate are ${\text{M}}_{\text{2}}\text{O}\text{and}{\text{M}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$, respectively.

Conclusion

The given reaction data equation, formulas for the oxide and Sulfate of the metal M was determined.