Question

A materials scientist has created an alloy containing aluminum, copper, and zinc, and wants to determine the percent composition of the alloy. The scientist takes a 11.470 g sample of the alloy and reacts it with concentrated HCl. The reaction converts all of the aluminum and zinc in the alloy to aluminum chloride and zinc chloride in addition to producing hydrogen gas. The copper does not react with the HCl. Upon completion of the reaction, a total of 9.77 L of hydrogen gas was collected at a pressure of 724 torr and a temperature of 27.0 °C. Additionally, 2.169 g of unreacted copper is recovered. Calculate the mass of hydrogen gas formed from the reaction.

- Calculate the mass of aluminum in the alloy sample

- what is the mass percent composition of the alloy (percentage of the Cu, Al and Zn)

Answer 1

PV= nRT

n H2=(724torr/760torr x1atm) x 9.77L/0.082L.atm/mol.Kx300K= 0.378341891 mol

mass H2= mol x MW = 0.378341891 mol x 2g/mol= 0.757g of H2

mass sample= mass Al + mass Zn + mass Cu

11.470g= mass Al + mass Zn + 2.169g

9.301=mass Al + mass Zn

3HCl + Al ----> AlCl3 + 3/2H2

2HCl + Zn ----> ZnCl2 + H2

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5HCl + Al + Zn -----> AlCl3 + ZnCl2 + 5/2H2

0.378341891 mol of H2 = 5/2 mol of H2

0.378341891 mol of H2 ---------- 5/2 mol H2

x=0.227005134 mol of H2 ------------- 3/2 molH2

0.227005134 mol of H2 produced at first reaction.

This calculation means that the total amount of moles calculated is the same as the 5/2 moles obtained by stoichiometry, so we need to calculate how much moles represent the 3/2 produced at the first reaction.

Let´s call X at the mass of Al, so, mass of Zn is 9.301 -X.

Mol Al= X/26.9815g/mol = 0.227005134 mol -----> X= 6.125g of Al

mass Zn= 9.301-6.125= 3.176g

%Cu= 18.9%

%Zn= 27.7%

%Al= 53.4%