Question

A mixture of aluminum and iron weighting 9.62g react with hydrogen chloride in aqueous solution according to the parallel reaction.

$2Al+6HCl\to 2AlC{l}_3+3H_2$

$Fe+2HCl\to FeC{l}_2+H_2$

A 0.738g quality of hydrogen is involved when the metal reacts completely. Calculate the mass of ions in the original mixture.

Short answer

The mass of Al and Fe present in the mixture is 5.14g Al and 4.48g Fe.

Step-by-step solution

Calculation of moles of Al and Fe

We are given with mixture of aluminum and iron which react with hydrogen chloride in an aqueous solution. We need to calculate the mass of the ion in the mixture involved.

The mass of the mixture of aluminum and iron in 9.62g. Let us consider x the mass of Al and y the mass of Fe present in the mixture. Hence

x + y = 9.62→(1)

The numbers of moles of aluminum and iron are calculated using their molar masses.

$$\begin{gathered} MoleofAl=xg.Al\times \frac{1molAl}{26.98g.Al}=\frac{x}{26.98}mol \\ MoleofFe=yg.Al\times \frac{1mol.Fe}{55.85g.Fe}=\frac{y}{55.85}mol \end{gathered}$$

Calculation of number of moles of hydrogen

When this mixture reacts with hydrogen chloride, 0.738g of hydrogen is involved.

The number of moles of hydrogen present in 0.738g is:

$0.738.H_2\times \frac{1mol.H_2}{2.016g.H_2}=0.366mol.H_2$

The reactions are:

$$\begin{gathered} 2Al+6HCl\to 2AlC{l}_3+3H_2 \\ Fe+2HCl\to FeC{l}_2+H_2 \end{gathered}$$

The number of moles of hydrogen produced from x g of Al and y g of Fe are:

$$\begin{gathered} \frac{x}{26.98}mol.Al\times \frac{3mol.H_2}{2mol.H_2}=0.0556x.mol.H_2 \\ \frac{y}{55.86}mol.Fe\times \frac{1mol.H_2}{1mol.Fe}=0.0179y.mol.H_2 \end{gathered}$$

Calculation of the masses of Al and Fe present in the mixture.

As all the metals in the mixture react completely, the moles of hydrogen formed from Al and Fe will be added up, which equals the number of moles of hydrogen present in 0.738g.

Therefore:

$0.0556x+0.0179y=0.366\to \left(2\right)$

Multiplying equation (1) with 0.0556 gives:

$0.0556x+0.556y=0.535\to \left(3\right)$

Subtracting equation (2) from equation (3) gives:

$$\begin{gathered} 0.0556x+0.0556y=0.535 \\ 0.0556x+0.0179y=0.366 \\ 0.377y=0.169 \\ y=4.48 \end{gathered}$$

Substituting the value of y in equation (1) gives:

$$\begin{gathered} x+4.48=9.62 \\ x=5.14 \end{gathered}$$

Hence, the masses of Al and Fe present in the mixture are 5.14g and 4.48g respectively.