IChO 1975 · Theory problem 2

Composition of an Al-Zn-Si-Cu alloy

An alloy prepared for experimental purposes contains aluminium, zinc, silicon, and copper. If 1000 mg of the alloy are dissolved in hydrochloric acid, 843 cm³ of hydrogen (0 °C, 101.325 kPa) are evolved and 170 mg of an undissolved residue remain. A sample of 500 mg of the alloy when reacted with a \(\ce{NaOH}\) solution produces 517 cm³ of hydrogen at the above conditions and in this case remains also an undissolved fraction.

Problem:

2.1 Calculate the composition of the alloy in % by mass.

Relative atomic masses:

\(A_{\mathrm r}(\ce{Al}) = 26.98;\qquad A_{\mathrm r}(\ce{Zn}) = 65.37;\qquad A_{\mathrm r}(\ce{Si}) = 28.09;\qquad A_{\mathrm r}(\ce{Cu}) = 63.55.\)

Official English text of the problem.

Answers

  • 2.1 The alloy contains approximately 61.8% aluminium, 21.2% zinc, 12.0% silicon, and 5.0% copper by mass.

Solution

2.1

Aluminium and zinc dissolve in hydrochloric acid and release hydrogen. Silicon and copper remain undissolved, so the 170 mg acid residue consists of these two elements.

The balanced net ionic reactions are

\[ \ce{2Al + 6H+ -> 2Al^3+ + 3H2}, \]

\[ \ce{Zn + 2H+ -> Zn^2+ + H2}. \]

The species are hydrogen ions [H+], aluminium(III) ions [Al+3], zinc(II) ions [Zn+2], and hydrogen [H][H].

In sodium hydroxide solution, aluminium and zinc dissolve because of their amphoteric behaviour; silicon also reacts, releasing hydrogen. Copper remains undissolved. The balanced equations are

\[ \ce{2Al + 2OH- + 6H2O -> 2[Al(OH)4]- + 3H2}, \]

\[ \ce{Zn + 2OH- + 2H2O -> [Zn(OH)4]^2- + H2}, \]

\[ \ce{Si + 2OH- + H2O -> SiO3^2- + 2H2}. \]

Here hydroxide is [OH-] and water is O. The products are tetrahydroxoaluminate(III), [Al-](O)(O)(O)O; tetrahydroxozincate(II), [Zn-2](O)(O)(O)O; and formally metasilicate, [O-][Si](=O)[O-]. The two hydroxo complexes have tetrahedral metal centres, which these SMILES do not specify. The silicon equation uses the conventional metasilicate representation; aqueous hydration and speciation do not change the hydrogen yield.

Thus, each mole of aluminium produces \(3/2\) mol of hydrogen in either medium, each mole of zinc produces \(1\) mol, and each mole of silicon produces \(2\) mol in alkali only.

Use a 1.000 g alloy basis and the molar gas volume at the stated conditions,

\[ V_{\mathrm m}=22.414\ \mathrm{dm^3\,mol^{-1}}. \]

The alkaline experiment, scaled from 500 mg to 1000 mg of alloy, would produce

\[ V_{\mathrm{alkali}}=2(517)=1034\ \mathrm{cm^3}. \]

Because aluminium and zinc produce the same amounts of hydrogen in both media, the excess is entirely attributable to silicon:

\[ \Delta V=1034-843=191\ \mathrm{cm^3} =0.191\ \mathrm{dm^3}. \]

Consequently,

\[ n_{\ce{Si}} =\frac{\Delta V}{2V_{\mathrm m}} =\frac{0.191}{2(22.414)} =4.26073\times10^{-3}\ \mathrm{mol}, \]

\[ m_{\ce{Si}} =n_{\ce{Si}}M_{\ce{Si}} =(4.26073\times10^{-3})(28.09) =0.119684\ \mathrm g. \]

The remainder of the acid residue is copper:

\[ m_{\ce{Cu}}=0.170-0.119684 =0.050316\ \mathrm g. \]

The total mass of aluminium and zinc is therefore

\[ m_{\ce{Al}}+m_{\ce{Zn}}=1.000-0.170=0.830\ \mathrm g. \]

The acid hydrogen yield gives a second equation:

\[ \frac32\frac{m_{\ce{Al}}}{26.98\ \mathrm{g\,mol^{-1}}} +\frac{m_{\ce{Zn}}}{65.37\ \mathrm{g\,mol^{-1}}} =\frac{0.843\ \mathrm{dm^3}} {22.414\ \mathrm{dm^3\,mol^{-1}}} =0.0376104\ \mathrm{mol}. \]

Substituting \(m_{\ce{Zn}}=0.830\ \mathrm g-m_{\ce{Al}}\) and solving gives

\[ m_{\ce{Al}}=0.618212\ \mathrm g, \qquad m_{\ce{Zn}}=0.211788\ \mathrm g. \]

Finally,

\[ w_i=100\frac{m_i}{1.000\ \mathrm g}, \]

so the composition, rounded to \(0.1\) percentage point, is

Element Percentage by mass
Aluminium 61.8%
Zinc 21.2%
Silicon 12.0%
Copper 5.0%
Total 100.0%

The solution was checked against the official solution.