Chapter 11, Problem 11.110QP

A 180.0-mg sample of an alloy of iron and metal X is treated with dilute sulfuric acid, liberating hydrogen and yielding Fe²⁺ and X³⁺ ions in solution. It is known that the alloy contains 20.0 percent iron by mass. The alloy yields 50.9 mL of hydrogen collected over water at 22°C and a total pressure of 750.0 torr. What is element X?

Interpretation Introduction

Interpretation:

The element X should be identified.

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.

$\text{PV = nRT}$

Use the ideal gas equation, $\text{PV = nRT}$ to calculate the number of moles of ${\text{H}}_{\text{2}}$ collected. By using the balanced equation, we can determine the amount of ${\text{H}}_{\text{2}}$ produced by the iron.  The remaining ${\text{H}}_{\text{2}}$ must have been produced from metal X.  Using the balanced equation, we can determine the number of moles of X that must have reacted to produce the remaining moles of ${\text{H}}_{\text{2}}$.

Answer to Problem 11.110QP

The element with a molar mass of $\text{157}\text{.2 g/mol is}\text{gadolinium,Gd}\text{.}$

Explanation of Solution

To calculate the total pressure of water and the pressure of Hydrogen in atm.

The partial pressure of water at $\text{22°C is 19}\text{.8 torr}\text{.}$  The total pressure is the sum of the partial pressures of the individual components in the mixture, in this case hydrogen and water. Therefore, $\begin{array}{l}{\text{P}}_{\text{total}}\text{=}{\text{P}}_{{\text{H}}_{\text{2}}}\text{+}{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\hfill \\ \hfill \\ \text{where}{\text{P}}_{\text{total}}\text{= 750}\text{.0 torr}\text{.}\hfill \end{array}$

Solving  for ${\text{P}}_{{\text{H}}_{\text{2}}}$ gives

$\begin{array}{l}{\text{P}}_{{\text{H}}_{\text{2}}}\text{=}{\text{P}}_{\text{total}}\text{-}{\text{P}}_{{\text{H}}_{\text{2}}\text{O}}\text{=750}\text{.0}\text{torr}\text{-}\text{19}\text{.8}\text{torr}\hfill \\ \text{=}\text{730}\text{.2}\text{torr}\hfill \\ \hfill \end{array}$

Converting to atm

$\begin{array}{l}\text{730}\text{.2}\text{torr}\text{×}\frac{\text{133}\text{.322}\text{Pa}}{\text{1}\text{torr}}\text{×}\frac{\text{1}\text{atm}}{\text{101,325Pa}}\text{=}\text{0}\text{.96079}\text{atm}\\ \text{T=295}\text{.15}\text{K}\text{(22°C}\text{+}\text{273}\text{.15}\text{)}\text{and}\text{V}\text{=}\text{0}\text{.0509}\text{L}\text{(50}\text{.9}\text{mL)}\end{array}$

Given that $\text{20\% of the 180}\text{.0 mg}$ alloy sample is iron by mass, the amount of iron in the alloy is $\text{36 mg or 0}\text{.036 g }\left(\text{0}\text{.20 × 180}\text{.0 mg}\right)\text{.}$ The difference between the mass of the alloy and the mass of Iron is the mass of metal X,

$\text{180}\text{.0 mg – 36 mg = 144 mg }\left(\text{0}\text{.144 g}\right)\text{.}$

The total pressure is the sum of the partial pressures of the individual components in the mixture, in this case hydrogen and water.

The Hydrogen pressure was calculated, the pressure of water is subtracts from total pressure then we get the Hydrogen pressure is $\text{=}\text{730}\text{.2}\text{torr}$

The pressures in atm are calculated by plugging in the values of the given 1 torr and 1 Pa.  The pressures in atm was found to be $\text{0}\text{.96079}\text{atm}$

To calculate the number of moles ${\text{H}}_{\text{2}}$ collected.

${\text{ n}}_{{\text{H}}_{\text{2}}}\text{=}\frac{\text{PV}}{\text{RT}}\text{= }\frac{\text{(0}\text{.96079}\text{atm}\text{)(0}\text{.0509}\text{L)}}{\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(295}\text{.15}\text{K)}}\text{=}\text{2}\text{.019}\text{×}{\text{10}}^{\text{-3}}\text{mol}{\text{H}}_{\text{2}}$

The moles of ${\text{H}}_{\text{2}}$ is calculated by plugging in the values of the given pressure, volume and temperature.  The moles of ${\text{H}}_{\text{2}}$ was found to be $\text{2}\text{.019}\text{×}{\text{10}}^{\text{-3}}\text{mol}{\text{H}}_{\text{2}}$

To calculate the number of moles ${\text{H}}_{\text{2}}$ collected.

${\text{ n}}_{{\text{H}}_{\text{2}}}\text{=}\frac{\text{PV}}{\text{RT}}\text{= }\frac{\text{(0}\text{.96079}\text{atm}\text{)(0}\text{.0509}\text{L)}}{\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(295}\text{.15}\text{K)}}\text{=}\text{2}\text{.019}\text{×}{\text{10}}^{\text{-3}}\text{mol}{\text{H}}_{\text{2}}$

The moles of ${\text{H}}_{\text{2}}$ is calculated by plugging in the values of the given pressure, volume and temperature.  The moles of ${\text{H}}_{\text{2}}$ was found to be $\text{2}\text{.019}\text{×}{\text{10}}^{\text{-3}}\text{mol}{\text{H}}_{\text{2}}$

Iron yields ${\text{Fe}}^{\text{2+}}$ upon reaction with dilute sulphuric acid. Writing a balanced equation for the reaction of iron with dilute Sulfuric acid gives:

${\text{Fe}}_{\left(\text{s}\right)}{\text{ + H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{aq}\right)}\to {\text{FeSO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{ + H}}_{\text{2}\left(\text{g}\right)}$

Next, determine the amount of ${\text{H}}_{\text{2}}$ produced by the iron using the mole ratio between Fe and ${\text{H}}_{\text{2}}$:

$\text{0}\text{.036}\text{g}\text{Fe}\text{×}\frac{\text{1}\text{mol}\text{Fe}}{\text{55}\text{.85}\text{g}}\text{×}\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}}{\text{1}\text{mol}\text{Fe}}\text{=}\text{6}\text{.446}\text{×}{\text{10}}^{\text{-4}}\text{mol}{\text{H}}_{\text{2}}$

The remaining moles of ${\text{H}}_{\text{2}}$ are produced from metal X:

$\text{2}{\text{.019 × 10}}^{\text{–3}}{\text{mol H}}_{\text{2}}\text{– 6}{\text{.446 × 10}}^{\text{–4}}{\text{mol H}}_{\text{2}}\text{= 1}{\text{.374 × 10}}^{\text{–3}}{\text{mol H}}_{\text{2}}$

Metal X yields ${\text{X}}^{\text{3+}}$ upon reaction with dilute Sulfuric acid. Writing the equation for the reaction of metal X with dilute Sulfuric acid gives:

${\text{X}}_{\left(\text{s}\right)}{\text{ + H}}_{\text{2}}{\text{SO}}_{\text{4}\left(\text{aq}\right)}\to {\text{X}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}{}_{\left(\text{aq}\right)}{\text{+ H}}_{\text{2}\left(\text{g}\right)}$

The balanced equation is:

${\text{2X}}_{\left(\text{s}\right)}{\text{ + 3H}}_{\text{2}}{\text{SO}}_{\text{4}\left(\text{aq}\right)}\to {\text{X}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}\left(\text{aq}\right)}{\text{ + 3H}}_{\text{2}\left(\text{g}\right)}$

Finally, using the mole ratio between X and H2, determine the number of moles of X needed to produce $\text{1}{\text{.374 × 10–3 mol H}}_{\text{2}}$.

$\text{1}\text{.374}\text{×}{\text{10}}^{\text{-3}}\text{mol}{\text{H}}_{\text{2}}\text{×}\frac{\text{2}\text{mol}\text{X}}{\text{3}\text{mol}{\text{H}}_{\text{2}}}\text{=}\text{9}\text{.16}\text{×}{\text{10}}^{\text{-4}}\text{mol}\text{X}$

Since there are 0.144 0g of metal X, the molar mass is:

$\frac{\text{0}\text{.144}\text{g}}{\text{9}\text{.16}\text{×}{\text{10}}^{\text{-4}}\text{mol}\text{X}}\text{=}\text{157}\text{.2}\text{g/mol}$

The element with a molar mass of $\text{157}\text{.2 g/mol isgadolinium,Gd}\text{.}$

Conclusion

The element X was identified.