Please Help. A materials scientist has created an alloy containing aluminum, cop
ID: 481548 • Letter: P
Question
Please Help.
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 12.302 g sample of the alloy and reacts it with concentrated HCI. 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 HCI. Upon completion of the reaction, a total of 11.2 L of hydrogen gas was collected at a pressure of 724 torr and a temperature of 27.0^degree C. Additionally, 2.277 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?Explanation / Answer
2Al+6HCl --à 2AlCl3 + 3H2 (1) and Zn+ 2HCl --à ZnCl2+ H2(2)
Mass of sample alloy = 12.302g
Mass of copper remaining = 2.277 gm
Mass of aluminimum and Zn= 12.302-2.277=10.025 gm
Let x= mass of Al and y= mass of Zn
Hence x+y =10.025 gm (3)
Hence x=10.025-y (3A)
Hydrogen gas is collected at P= 724 Torr = 724/760 atm =0.953 atm, V= 11.2 L, T= 27 deg.c =27+273=300 K
R =0.0821 L.atm/mole.K, n= PV/RT = 0.953*11.2/(0.0821*300)=0.43 mole
Mass of hydrogen = moles* molecular weight =0.43*2= 0.86 gm
Atomic weights : Al =27 and Zinc = 65.4
Moles of Al in the mixture = x/27 and that of Zinc = y/65.4
From reaction (1), 2 moles of Al produces 3 moles of Al
x/27 moles produces x*3/54 =0.0556x
from reaction (2), 1 mole of Zinc produces 1 mole of H2
y/65.4 mole of Zinc produces y/65.4 moles of H2 =0.0153 moles of H2
total moles of H2, 0.0556x+0.0153y= 0.43 (4)
From 3A, 0.0556*(10.025-y)+0.0153y= 0.43
0.56-0.0556y+0.0153y =0.43
Hence y*(0.0556-0.0153)= 0.56-0.43 =0.13
Hence y (mass of zinc)= 3.23 gm and x ( mass of Aluminium)= 10.025-3.23=6.795 gm
Mass % : Zinc = 100*3.23/12.302=26.26%, Aluminium = 100*6.795/12.302=55.23%, Copper = 100-(55.23+26.26) =18.51%