Posted by Jeremiah on Monday, November 19, 2012 at 6:29pm.
472 mL of H2 gas was collected over water when 1.256 g of Zn reacted with excess HCl. The atmospheric pressure during the experiment was 754 mm Hg and the temperature was 26 degrees C.
A. Write the balanced chemical equation (I have this)
Zn + HCl ---> ZnCl2 + H2
B. What is the water vapor pressure at 26 degrees C? Using the chart provided I have 25.2
C. What is the partial pressure (in atmospheres) of dry hydrogen gas in the mixture?
Not sure if I have this correct, but I came up with: 0.959 atm
That's right.
D. Calculate the number of moles of H2 produced by this reaction using the ideal gas law.
I have 0.0184 mol H2
I have (0.959x0.427)/(0.08206*299)] = n = 0.01669 which you can round.
E. Use the data from the experiment to calculate the experimental molar mass of zinc in g/mol.
I think I have this part correct. 0.0192 g/mol
mol = g/molar mass; rearrange to molar mass = (g/mol) = 1.256/0.01669 = 75.26 g/mol
F. What is the molar mass of zinc from the periodic table (known value) 65.39 g/mol
G Calculate the percent error for the experimental molar mass of zinc.
I am confused here. Using the values I have above I am coming up with 99.97%.
% error = [(exp value-actual value)/(actual value)]*100 =
[(75.26-65.38)/65.38]*100 = about 15% but that's just a close number.
I went to our school's tutoring lab and received the explanation I needed to understand this. However I have another related question. I don't know if you will see this or not. Suppose we have a similar problem, except using Aluminum and HCl. The ratio is different. I am lost on part E again. In the above problem I understand why you can figure out the molar mass of zinc because 1 mole zinc is consumed for every 1 mole H2 produced. In the equation 2Al + 6HCl ---> 2AlCl3 + 3H2 for every 2 Al consumed 3 moles of H2 is produced. I am having a problem understanding how you would find the molar mass of Aluminum.
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