Under the conditions used in your laboratory experiment, 0.0513 grams of aluminu
ID: 587557 • Letter: U
Question
Under the conditions used in your laboratory experiment, 0.0513 grams of aluminum (At. Wt. 27.0) was eacted with excess hydrochloric acid (HCI) to produce hydrogen gas (H2) according to the following palanced equation: Determine the number of moles of H2 evolved from that quantity of aluminum when collected over water. 4 points If the volume of gas generated was 75.0 mL and the temperature was 27.0 °C, determine the total pressure of the hydrogen gas evolved from the quantity of aluminum stated above. [5 points) The vapor pressure of water at 27.0 °C was 26.7 mmHg, determine the partial pressure of the hydrogen gas collected in mmHg. (4 points)Explanation / Answer
2Al + 6HCl -------------> 2AlCl3 + 3H2
no of moles of Al = w/G.M.Wt
= 0.0513/27 = 0.0019 moles
2moles of Al react with HCl to gives 3 moles of H2
0.0019 moles of Al react with HCl to gives = 3*0.0019/2 = 0.00285 moles
PV = nRT
V = 75ml = 0.075 L
T = 27+273 = 300K
P = nRT/V
= 0.00285*0.0821*300/0.075 = 0.936atm
total pressure = 0.936atm = 0.936*760 mmHg = 711.36mmHg
vapor pressure of water at 27C^0 = 26.7mmHg
The pressure of H2 = total pressure - vpor pressure of water
= 711.36-26.7 = 684.66mmHg
partial pressure of H2 = 684.66mmHg
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