A materials scientist has created an alloy containing aluminum, copper, and zinc
ID: 571906 • Letter: A
Question
A materials scientist has created an alloy containing aluminum, copper, and zinc, and wants to determine the percent composition of the alloy. Thé scientist takes a 11.641 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 8.91 L of hydrogen gas was collected at a pressure of 98.2 kPa and a temperature of 27.0 °C. Additionally, 2.966 g of unreacted copper is recovered Calculate the mass of hydrogen gas formed from the reaction. Number Calculate the mass of aluminum in the alloy sample. Number What is the mass percent composition of the alloy? Number Number Number % Cu %AlExplanation / Answer
Ans. Step 1: Given,
Volume of H2 gas collected = 8.91 L
Temperature, T = 27.00C = 300.150C
Pressure, P = 98.2 kPa = 0.969 atm
Using Ideal gas equation: PV = nRT - equation 1
Where, P = pressure in atm
V = volume in L
n = number of moles
R = universal gas constant= 0.0821 atm L mol-1K-1
T = absolute temperature (in K) = (0C + 273.15) K
Putting the values in above equation-
0.969 atm x 8.91 = n x (0.0821 atm L mol-1K-1) x 300.15 K
Or, n = 8.63379 atm L / (24.642315 atm L mol-1)
Hence, n = 0.350 mol
Therefore, moles of H2 collected = 0.350 mol
Mass of H2 collected = Moles x Molar mass
= 0.350 mol x (2.01588 g / mol)
= 0.705558 g
# Step 2: Combined mass of (Zn + Al) = Total alloy mass – Mass of Cu
= 11.641 g – 2.966 g
= 8.675 g
# Step 3: Let the mass of Al in alloy = X gram
Mass of Zn in alloy = Y gram.
Now, the sum of masses of Zn and Al is equal to 8.675 g.
So,
X g + Y g = 8.675 g
Or, X + Y = 8.675 - equation 1
# Step 4: Balanced reactions: 2 Al(s) + 6 HCl(aq) ------------> 2 AlCl3 + 3 H2(g)
Zn(s) + 2 HCl(aq) -------------> ZnCl2 + H2(g)
According to the stoichiometry of balanced reaction, 2 mol Al produces 3 mol H2 ; and 1 mol Zn produces 1 mol H2.
# Moles of Al in alloy = Mass / Molar mass = X g / (26.981 g/mol) = 0.03706X mol
Moles of H2 produced by Al = (3 mol H2 / 2 mol Al) x 0.03706X mol
= 0.05559X mol H2
# Moles of Zn in alloy = Mass / Molar mass = Y g / (65.39 g/mol) = 0.01529Y mol
Moles of H2 produced by Zn = (1 mol H2 / 1 mol Zn) x 0.01529Y mol Zn
= 0.01529Y mol H2
# The sum of total moles of H2 produced must be equal to 0.350 mol (as calculated in step 1).
So,
0.05559X mol + 0.01529Y mol = 0.350 mol
Hence, 0.05559X + 0.01529Y = 0.350 - equation 2
# Step 4: Comparing (equation 1 x 0.05559) – equation 2
0.05559X + 0.05559Y = 0.48224325
(-) 0.05559X + 0.01529Y = 0.350
----------------------------------------------
0.0403Y= 0.13224325
Or, Y = 0.13224325 / 0.0250
Hence, Y = 3.2815
Therefore, mass of Zn in alloy = Y g = 3.2815 g
# Putting the values in equation 1-
X = 8.675 g - 3.2815 g
Hence, Y = 3.38527 g
Therefore, mass of Al in alloy = X g = 5.3935 g
# Step 5: % Cu in alloy = (Mass of Cu / Mass of alloy) x 100
= (2.966 g / 11.641 g) x 100
= 25.479 %
% Al in alloy = (5.3935 / 11.641) x 100 = 46.332 %
% Zn in alloy = (3.2815/ 11.641) x 100 = 28.189 %
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