PROP1003 Production of Hydrogen Gas 129 Post-Lab Questions 1.Show by calculation

Question

PROP1003 Production of Hydrogen Gas 129 Post-Lab Questions 1.Show by calculation that 20 mL of concentrated HCI is more than enough to react your mass of magnesium. ("Concentrated" HCI is approximately 12 M) 2. A 1.73 gram sample of an unknown metal releases 380 ml. of hydrogen gas when collected over water at 20'C and 720 mm Hg (total buret pressure). Assuming the metal follows the same stolch- iometry as magnesium, what is its atomic mass? What is its identity? 3. How many grams of zinc would be required to produce 15.0 L of hydrogen gas at 1.08 atm anda temperature of 25.0°C? Assume excess hydrochloric acid. PT 142 CO 032 Why must the rubber stopper that holds the copper wire in place have a hole in it? 2 3

Explanation / Answer

1. moles of HCl used = molarity x volume

= 12 M x 0.02 L = 0.24 mol

This amount of HCl has reacted completely with the metal present as moles of metal used was lower than the moles of HCl added. HCl is the reagent present in excess in the reaction.

2. Volume of H2 gas (V) = 380 ml = 0.38 L

Temperature (T) = 20 oC + 273 = 293 K

Vapor pressure of H2O at 20 oC = 17.5 mmHg

Pressure of H2 collected = 720 - 17.5 = 702.5 mmHg

Pressure (P) = 702.5/760 = 0.924 atm

R = gas constant

So,

moles of H2 gas collected = PV/RT

= 0.924 x 0.380/0.0821 x 293

= 0.0146 mol

moles of metal = 0.0146 g

molar mass of metal = 1.73 g/0.0146 mol = 118.523 g/mol

Identity of metal = Sn (tin)

3. moles of Zn to be reacted = PV/RT

P = 1.08 atm ; V = 15.0 L ; R = gas constant ; T = 25 oC + 273 = 298 K

So,

moles of Zn needed = 1.08 x 15/0.0821 x 298 = 0.662 mol

mass of Zn needed to produce 15.0 L H2 gas = 0.662 mol x 65.38 g/mol = 43.29 g

4. The hole in the stopper having the copper wire is used as a vent for the closed system.

5. The standard reduction potential of Al3+/Al and Mg2+/Mg are lower than the H+/H2 system, that is the two metals would act as reducing agent in the given reaction with HCl. This produces Al3+ and Mg2+ in the system along with H2 gas readily. Whereas, for Cu, the reduction potential is higher than for hydrogen and thus it acts as an oxidizing agent instead, so the reaction is not feasible with HCl.

6. Chemical equation,

Mg(s) + 2HCl(aq) ---> MgCl2(aq) + H2(g)

Species oxidized = Mg

Species reduced = HCl

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