in gas law analysis of an Aluminium-Zinc Alloy experiment : 1- Discuss the error
ID: 1059380 • Letter: I
Question
in gas law analysis of an Aluminium-Zinc Alloy experiment :
1- Discuss the error (high or low) for the calculated %Al if the barometric pressure is incorrectly recorded as 750.0 cm instead of 75.00cm. Would the answer itself have altered you to the fact that an error had been made ?
2- Disccus the error (high or low) on the %Al resulting from forgetting to consider the vapor pressure of the water in the calculation.
3- the reaction of the alloy with the acid is exothermic. if the hydrogen gas released is Warner than the water temperature and you use the incorrect low temperature, how would this effect your calculated %Al result (high or low) explain briefly
please answer me
Inility 200 Tuol Tube. Ring Stand Rubber Tabing 250- 400 mL BenkerExplanation / Answer
1. The reactions taking place are:
2 Al (s) + 6 H+ (aq) -----> 2 Al3+ (aq) + 3 H2 (g) …..(1)
Zn (s) + 2 H+ (aq) -----> Zn2+ (aq) + H2 (g) ……(2)
The experiment requires us to calculate the moles of H2 produced by using the ideal gas law:
PH2.V = nH2.R.T where T = reaction temperature.
When the correct barometric pressure is 75.00 cm (=750 mm = 750mm/760mm*1 atm = 0.9868 atm), we have,
nH2 (1) = 0.9868.V/RT
When the barometric pressure is recorded as 750.00 cm (= 7500 mm = 7500mm/760mm*1atm = 9.868 atm), we have
nH2 (2) = 9.868.V/RT
Since V is directly recorded (by water displacement) and R and T are constants, nH2 (2) = 10*nH1 (1).
The number of moles of H2 liberated is calculated to be much higher. Now, as per equations (1) and (2), both the reactions liberate H2 and the sum total of the number of moles of H2 is nH2.
Since nH2 is calculated to be higher, nH2 (Al) and nH2 (Zn) will be individually higher; as nH2 is directly proportional to the masses of Al and Zn taken, the calculated masses will be higher and hence the sum total will be higher. However, the percentage of Al in the alloy is given by
%Al = (mass of Al/(mass of Al + mass of Zn)*100.
As both the numerator and the denominator increase proportionately (by noting a wrong pressure and hence a wrong nH2), the percentage Al will remain more or less similar to the actual value. There will be no appreciable difference in calculating %Al (ans).
2. We know that barometric pressure, P = PH2 + Pwater.
If the student forgets to consider the vapour pressure, then he calculates PH2 = P which is higher than the actual pressure of dry hydrogen gas.
Since PH2 is directly proportional to nH2 (as we showed above), we will calculate a higher nH2 and hence a higher mass of Al. However, the same error will manifest itself in the calculation of Zn so that the sum total of the masses of the metals increase. The percentage of Al will remain the same as before, because both mass Al and mass total are proportionately higher than the actual value (ans).
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