1. Calculate n(H2) obtained from 1.000 gram of Al. Also, calculate n(H2) obtaine
ID: 905214 • Letter: 1
Question
1. Calculate n(H2) obtained from 1.000 gram of Al. Also, calculate n(H2) obtained from 1.000 gram of Zn.
2. Prepare a graph for a 1.000 gram sample of alloy with n(H2) on the vertical axis and percent Al on the horizontal axis. The graph will be linear. Plot four points, one where the percent Al is 100 (i.e., 1.000 gram of Al) and another point where the percent Al is zero (1.000 gram of zinc). Choose any two other percentages for a 1.000 gram sample, calculate the moles of hydrogen that will be obtained, and plot the points.
Explanation / Answer
1)
-> For Aluminum :
Al + 3HCl -----> AlCl3 + 1.5 H2
Thus each atom of aluminium releases 1.5 moles of H2 gas. So, 1 mole of aluminum releases 1.5 moles of H2 gas. The number of moles of Aluminum involved = 1/27 moles
So the number of H2 released = 1.5/27 = 1/18 moles = 0.0555556 moles
->For zinc
Zn + 2HCl ----> ZnCl2 + H2
Thus each atom releases 1 molecule of H2. The number of moles of Zn involved = 1/65.4 = 0.0153 moles
So, the number of moles of H2 = 0.0153 moles
n(H2) in case of Aluminum = 0.055556
n(H2) in case of Zinc = 0.0153
2)
The reaction is written as :
Zn + 2HCl ------> ZnCl2 + H2
Thus each mole of Zn releases 1 mole of H2. Number of moles of Zn = 1/65.4 =1.529 * 10^-2 moles
At x=0, y = 1.529*10^-2
The point corresponding to the 100 on x means that we have 1 gram of aluminum.
The reaction is written as :
Al + 3HCl ------> AlCl3 + 1.5H2
Thus each mole of Al releases 1.5 mole of H2. Number of moles of Al = 1/27 =3.7037* 10^-2 moles
n(H2) = 5.555* 10^-2 moles
So at x = 100, y = 5.555*10^-2 moles
So the graph of the equation is written as
(y-1.529*10^-2) = 0.0402/100x
y-(1.529*10^-2) = (4.02 * 10^-4) x
Slope is 4.02*10^-4
Y intercept is 1.529*10^-2 moles
An alloy is a mixture of 2 similar non reacting elements, usually metals.
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