Waste Oispose of all waste according to the direction of the InstructU 5o-mi eud

Question

Waste Oispose of all waste according to the direction of the InstructU 5o-mi eudiometer funnel Materials 250-mL wash bottle 250-mi beaker Copper wire (20 cm) 10-ml graduated cylinder Sandpaper Magnesium ribbon (4-5 cm) Metric ruler Pre Laboratory Assignment Answer the following on o separote sheet of paper Define the following terms Eudiometer Single-displacement reaction Atmospheric pressure Activity Complete the following 1. Assume a small piece of magnesium with mass the base of a eudiometer. The hydrogen gas generated in the reaction occupies a volume of 46.8 f 0.0461 g is reacted with hydrochloric acid at ml and leaves behind a column of acid that is 2.65 cm in height. Write a net-ionic equation for the reaction. b. If barometric pressure is measured to be 0.9952 atm, determine the pressure (in atm) of a. hydrogen gas generated assuming the temperature of the solution is 21°C. Determine the mole amount of hydrogen gas generated. d. c. Determine the mole amount of magnesium that reacted to generate the hydrogen gas. Assuming the mass of the magnesium was 0.0461 g, determine the molar of magnesium e. f. Given an actual molar mass of 24.31 g/mol, determine the percent error of the experimentally determined molar mass.

Explanation / Answer

Ans 1

Part a

The balanced reaction

Mg(s) + 2HCl (aq) = MgCl2 (aq) + H2(g)

Molecular ionic equation

Mg(s) + 2H+ (aq) + 2Cl- (aq) = Mg2+(aq) + 2Cl-(aq) + H2

Net ionic equation

Mg(s) + 2H+(aq) = Mg2+(aq) + H2

Part b

Barometric pressure = 0.9952 atm

Vapor pressure of water at 21°C

= 18.7 mmHg x 1atm/760mmHg

= 0.024605 atm

P water column = 2.65cm x (10^-2cm/10^-3cm) x 0.997/13.6

= 1.9426 mmHg x 1atm/760mmHg

= 0.002556 atm

Pressure of H2 gas generated = Barometric pressure - Vapor pressure of water - P water column

= 0.9952 - 0.024605 - 0.002556

= 0.968039 atm

Part c

From the ideal gas equation

Pressure x volume = moles x R x Temperature

Moles of H2 gas produced

= (0.968039 atm) x (46.8 mL x 1L/1000 mL) / (0.08206 L-atm/mol-K x 294K)

= 0.0018778 mol

Part d

Moles of Mg reacted = Moles of H2 gas produced

= 0.0018778 mol

Part e

Molar mass of Mg = mass/moles

= (0.0461 g) / (0.0018778 mol)

= 24.549 g/mol

Part F

% error = (experiment value - actual value) * 100 / (actual value)

= (24.549 - 24.31)*100/(24.31)

= 0.985 %

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