Question

Aluminum reacts with sulfuric acid to form aluminum sulfate and hydrogen. What is the volume of hydrogen gas in liters (L) produced at $300\mathrm{K}$ and $1.0\mathrm{atm}$ pressure, when$5.4\mathrm{g}$of aluminum and $50.0\mathrm{mL}$ of $5.0\mathrm{M}$ sulfuric acid are combined for the reaction?

Answer 1

Given reaction:

$\underset{\left(\mathrm{Aluminium}\right)}{\underbrace{2\mathrm{Al}\left(\mathrm{s}\right)}}+\underset{(\mathrm{Sulphuric}\mathrm{acid})}{\underbrace{3{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}}\to \underset{(\mathrm{Aluminium}\mathrm{sulphate})}{\underbrace{{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\left(\mathrm{s}\right)}}+\underset{(\mathrm{Hydrogen}\mathrm{gas})}{\underbrace{3{\mathrm{H}}_2\left(\mathrm{g}\right)}\uparrow }$

Given:

$$\begin{gathered} \mathrm{T}=300\mathrm{K} \\ \mathrm{P}=1.0\mathrm{atm} \\ {\mathrm{V}}_{\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)}=50.0\mathrm{mL} \\ {\mathrm{Molarity}}_{\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)}=5.0\mathrm{M} \end{gathered}$$

Given mass of $\mathrm{Al}=5.4\mathrm{g}$

Molar mass of $\mathrm{Al}=27\mathrm{g}$

Step 1: Find the number of moles of Aluminium and Sulphuric acid.

We know that,

$$\begin{gathered} \mathrm{No}.\mathrm{of}\mathrm{moles}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}} \\ \mathrm{No}.\mathrm{of}\mathrm{moles}=\frac{5.4\mathrm{g}}{27{\mathrm{gmol}}^{-1}} \\ \mathrm{No}.\mathrm{of}\mathrm{moles}=0.2\mathrm{moles} \end{gathered}$$

Similarly, we could find out the number of moles of Sulphuric acid.

$$\begin{gathered} \mathrm{No}.\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{H}}_2{\mathrm{SO}}_4=\frac{\mathrm{M}\times \mathrm{V}}{1000} \\ \mathrm{No}.\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{H}}_2{\mathrm{SO}}_4=\frac{5\times 50}{1000}=0.25\mathrm{moles} \end{gathered}$$

$\mathrm{As},\frac{0.2}{2}>\frac{0.25}{3},{\mathrm{H}}_2{\mathrm{SO}}_4$is the limiting reagent (The substance that is totally consumed in a reaction is called the limiting reagent).

Now by using the stoichiometric concept (Stoichiometry as the calculation of products and reactants in a chemical reaction. It is basically concerned with numbers.), one could say,

When 2 moles of Aluminum are made to react with 3 moles of Sulphuric acid, 3 moles of hydrogen gas are formed.

$\mathrm{Moles}\mathrm{of}\mathrm{hydrogen}\mathrm{formed}=\frac{3\times 0.25}{3}=0.25\mathrm{moles}.$

Step 2: Find the volume of hydrogen gas.

Now by Idea gas equation,

$\mathrm{PV}=\mathrm{nRT}$

Where,

$\mathrm{P}=$Pressure,

$\mathrm{V}=$Volume

$\mathrm{n}=$no. of moles

$\mathrm{R}=0.082{\mathrm{LatmK}}^{-1}{\mathrm{mol}}^{-1}$ (gas constant)

$\mathrm{T}=$Temperature.

$$\begin{gathered} \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}} \\ \mathrm{V}=\frac{0.25\times 0.082\times 300}{1} \\ \mathrm{V}=6.15\mathrm{L} \end{gathered}$$

Therefore the amount of hydrogen gas produced is $6.15\mathrm{L}$