Equivalent Weight of Magnesium

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Equivalent Weight of Magnesium Experiment # 16

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Equivalent Weight of Magnesium. Experiment # 16. What we are doing today:. We are going to verify the atomic mass of Magnesium. Mg =24.12 g We are going to determine the # of moles of hydrogen displaced during a reaction. Going to use the Ideal Gas Law. - PowerPoint PPT Presentation

Transcript of Equivalent Weight of Magnesium

Page 1: Equivalent Weight of Magnesium

Equivalent Weight of Magnesium

Experiment # 16

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What we are doing today:

• We are going to verify the atomic mass of Magnesium. •Mg =24.12 g

• We are going to determine the # of moles of hydrogen displaced during a reaction.

Going to use the Ideal Gas Law. •Once we calculate the # of moles of Hydrogen produced we can calculate the # of moles of Mg that reacted.

Going to use Stoiciometry and dimensional analysis

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What we are doing today:

We will then verify the atomic mass of magnesium by comparing the grams of Mg used to the number of moles of Mg used.

Grams of Mg / Moles of Mg

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Idea Gas Law PV=NRT

P pressure in atmospheres (atms)

V volume in liters (l)

N number of moles

R gas constant = 0.0821 L atm /mol K

T temperature in Kelvin

In order to use the Ideal gas law the following information must be true:

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Atm 1 atm = 760 mmHg

L 1000ml = 1L

T K = 273 + OC

Need to convert:

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Mg + HCl → H2 + MgCl2

Procedures

• Obtain approximately 5.0 cm piece of Mg. and scrape with steel wool

• Using a mg balance accurately determine the mass of the Mg.

• Coil the Mg around a pencil

• Wrap a 12cm string around the coil• Leave aproximate 8cm as a handle

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Mg + HCl → H2 + MgCl2

Procedures

• Obtain a 1000 mL. beaker and fill it with water

• Add 10ml of 6M HCl to your eudiometer tube.

• With a 150 ml Beaker slowly fill the tube with water.

• DO NOT MIX THE WATER AND THE ACID.

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Procedures Continued

Carefully lower the coil into the tube holding the string.

• Use your thumb to close the tube and then invert it.

• Place your tube in your 1000 mL. beaker and clamp it in place.

1. The mouth of the eudiometer tube should be resting on the bottom of the 1000ml Beaker2. The Mg should be trapped in the eudiometer tube

3. The Mg should be reacting to the acid producing H2 gas.

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• Make the gas pressure inside the tube equal to the room pressure BEFORE measuring the volume. – This will make your water column measurement =0

• You do this by making the water level INSIDE the tube equal to the water level in the battery jar.

• You will raise or lower the tube to do this. You may have to add some water to the battery jar.

• . Be careful: do not raise the gas tube so high that its open end comes above the battery jar's water level.

Water column measurement

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Measuring the volume of gas produced

• When the gas pressures are equalized, read the volume of gas from the scale on the tube.

• Read it to the closest 0.1 mL and record this value in your data table.

• Please note that zero is at the top of the tube. • Read the value from the top down, not from the 50

mL mark going up.

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Measurements taken

•Measure the amount of gas produced based on water displacement

• Record the barometric pressure and convert to atm

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Use my data as trial # 1 and you do two trials

#1 = Mg mass = 0.040g #2 = Volume of H 2 collected in ml = 47.0 ml#3 =Volume of H 2 collected in (L) = 0.047L from eudiometer tube#4 = Barometric pressure: 759 mm of Hg. #5 = Water temp: 20 OC K = 273 + OC = 293 K#6 & 7 = 0

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#8 = Water vapor pressure

(Pressure of H2O vapor) see chart on pg 172 of lab book.

My water temp was 20OC and my pressure is 17.5 mm of Hg

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#9 = Pressure of dry H2 (g)

Total pressure = P1 +P2

Total pressure = atmospheric pressureP1 = pressure of water vaporP2 = pressure of Dry H2 gas

atmospheric pressure – water vapor pressure = Dry H2 gas

759 mm of Hg – 17.5mm of Hg = 741.50 mm of Hg Convert to atm = 0.98 atms

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#10 = moles of H2 (g)

NNRTPV RT

PV

0.0821 atms mol/k * 293K

0.98 atms * 0.047L = 0.002 mols or

2.0 x 10 –3 mols

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# 13 Mg/mol Mg

20.00 g/mole was the mass of Mg that I determined. The actual mass of Mg is 24.3 g/mole

Mg + HCl → H2 + MgCl2

Based on the balanced formula, For every 1 mol of H2 gas = 1 mol of Mg

Grams Mg/ mol H2 gas = Grams Mg/ mol mg = atomic mass of Magnesium

0.040g/ 2.0 x 10 –3 mols = 20g/mol

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Due next week

• Pg 174 and 175

• Need to show all work for full credit!!