Question

A mixture of aluminium and zinc weighing 1.67 grams was completely dissolved in acid and 1.69 litres of hydrogen gas was evolved. The hydrogen gas was measured at STP conditions. What was the mass percent of aluminium in the original mixture?
(A) 25
(B) 40
(C) 60
(D) 75

Answer 1

Hint: To solve the given problem, we need to have the basic knowledge of concepts like mole ratio, volume of molecules at STP, mass percent, etc. Knowing all these terms will help us get nearer the answer.
- Also, out of all the given options only one will be the answer.

Complete Solution :
Let us know about some terms before directly solving the given illustration,
-Atomic mass - It is the empirical mass of an element from the modern periodic table.
-It is empirical that one mole of atoms or molecules will have a volume of 22.4 L.
-The number of moles of the molecule is the ratio of the given mass to the molar mass.
-The mass percent is the quantity to measure concentration of a component in the mixture of components.

Now, let us move towards the illustration:
Let x gm of Al be present in the sample. Thus, (1.67-x) gm of Zn will be present in the same.
We know that,
The atomic mass of Al = 27 gm/mol
The atomic mass of Zn = 65.4 gm/mol
Thus,
The number of moles of Al present in the sample = $\dfrac{x}{27}moles$.
The number of moles of Zn present in the sample = $\dfrac{\left( 1.67-x \right)}{65.4}moles$.

According to the given illustration,
The mixture of Al and Zn was dissolved in acid resulting in the evolution of hydrogen gas. This can be shown as:
\[\begin{align}
 & 2Al+6{{H}^{+}}\to 2A{{l}^{3+}}+3{{H}_{2}} \\
 & Zn+2{{H}^{+}}\to Z{{n}^{2+}}+{{H}_{2}} \\
\end{align}\]

Now, considering stoichiometry we can say that,
Volume of hydrogen obtained from Al = $\dfrac{3\times 22.4\times x}{2\times 27}$
Volume of hydrogen obtained from Zn = $\dfrac{22.4\times \left( 1.67-x \right)}{65.4}$
as, one mole of any gas occupies a volume of 22.4 L.
Thus,
$\dfrac{3\times 22.4\times x}{2\times 27}$ + $\dfrac{22.4\times \left( 1.67-x \right)}{65.4}$ = 1.69
On solving this we get,
x = 1.248 gm.
- Hence, the mass of Al present in the sample is 1.248 gm.
- Thus, the percent of Al in the sample is given as,
- Mass percent of Al = $\dfrac{1.248}{1.67}\times 100 = 75%$
So, the correct answer is “Option D”.

Note: Do note to specify and use correct units while solving the problems as above. If the question was about the mass percent of Zn, then by the same process we can also determine the mass % of zinc.