The experiment involves the determination of the molar mass of aluminum by measu

Question

The experiment involves the determination of the molar mass of aluminum by measuring the amount of hydrogen gas liberated when a sample of aluminum is treated with excess hydrochloric acid. 1.)In the reaction between aluminum and hydrochloric acid, how many moles of hydrogen gas would be liberated when 3.437E-2 grams of aluminum are treated with excess hydrochloric acid? Moles of hydrogen gas liberated = ________ mol 2.)The atmospheric pressure was measured to be 0.9626 atm and the room temperature was 21.00oC. At this temperature, the vapor pressure of water is 18.60 torr. What would be the volume of the number of moles of hydrogen just calculated under these conditions? Enter the answer in mL. Volume of hydrogen gas = _________mL 3.)The syringe used in this experiment is calibrated to 60 mL. However, there is a 'dead volume' at the needle end of the syringe that is uncalibrated. This volume equals 1.20 mL. Bearing in mind that the syringe is upside down, what would the reading on the syringe be that corresponded to the volume you have just calculated? Reading on the syringe = ___________mL 4.)Suppose that you have completed this week's experiment and have determined that the molar mass of aluminum is 24.79 g mol-1. What is the percent error of your experimental result? Percent error of the experimental result = ________%

Explanation / Answer

Part I:
Given amount divided by 17.98733605
In this case: [(3.601E-2)/17.98733605] = 2.002E-3


Part II:
First, convert vapor pressure of water from torr to atm by dividing by 760.
(22.40)/760 = 2.947E-2 atm

Next, subtract this amount from the atmospheric pressure to find pressure you need.
.9543-(2.947E-2) = .9248

Convert temperature from celcius to kelvin by adding 273.
24+273 = 297

Use PV=nRT. Since you need volume it will be V=(nRT)/P
[(2.002E-3)(.0821)(297)]/.9248 = 5.279E-2 L

Convert to mL by multiplying by 1000.
(5.279E-2)1000 = 52.79 mL


Part III:
Subtract 1.2 from amount in part II.
52.79-1.2 = 51.59 mL


Part IV:
Divide given by 26.98.
24.99/26.98 = 9.262E-1

Multiply it by 100 to convert to percentage.
(9.262E-1)100= 92.62

Subtract this amount from 100.
100-92.62 = 7.38%

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