How many Litre of hydrogen at STP will be liberated when 1g of zinc reacts with excess of HCl?

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Apparatus notes
Health, safety and technical notes
Teaching notes

Magnesium reacts with dilute hydrochloric acid in a conical flask which is connected to an inverted measuring cylinder in a trough of water. The volume of hydrogen gas produced is measured over a few minutes, and the results are used to plot a graph

This is intended as a class practical. It is best if the students work in pairs because setting up and starting the experiment requires more than one pair of hands. One student can add the magnesium ribbon to the acid and stopper the flask, while the other starts the stopclock. During the experiment, one student can take the readings while the other records them. The experiment itself takes only a few minutes. But allow at least 30 minutes to give students time to set up, take readings and draw graph.

Hydrogen gas (extremely flammable) is generated in the experiment. Students should not have access to any source of ignition.

Equipment

Apparatus

Eye protection
Conical flask (100 cm3)
Single-holed rubber bung and delivery tube to fit conical flask (note 1)
Trough or plastic washing-up bowl (note 2)
Measuring cylinders (100 cm3), x2
Clamp stand, boss and clamp
Stopwatch
Graph paper

Apparatus notes

The bungs in the flasks need to be rubber. Corks are too porous and will leak. The tube through the bung should be a short section of glass, and then a flexible rubber tube can be connected.
Gas syringes can be used instead of troughs of water and measuring cylinders. But these are very expensive and are probably best used by the teacher in a demonstration. Syringes should not be allowed to become wet, or the plungers will stick inside the barrels.

Chemicals

Magnesium ribbon cut into 3 cm lengths
Dilute hydrochloric acid, 1M

Health, safety and technical notes

Read our standard health and safety guidance
Wear eye protection throughout. Ensure that there are no naked flames.
Magnesium ribbon, Mg(s) – see CLEAPSS Hazcard HC59a. The magnesium ribbon should be clean and free from obvious corrosion or oxidation. Clean if necessary by rubbing lengths of the ribbon with fine sandpaper to remove the layer of oxidation.
Hydrochloric acid, HCl(aq) – see CLEAPSS Hazcard HC47a and CLEAPSS Recipe Book RB043. The hydrochloric acid should be about 1M for a reasonable rate of reaction. Each experiment run will need 50 cm3. Though low hazard, eye protection is necessary as you may get a spray as tiny bubbles burst.
Hydrogen gas, H2(g) (EXTREMELY FLAMMABLE) – see CLEAPSS Hazcard HC048. Ensure that all naked flames are extinguished, and that there are no other sources of ignition available to students.

Procedure

Measure 50 cm3 of 1M hydrochloric acid using one of the measuring cylinders. Pour the acid into the 100 cm3 conical flask.
Set up the apparatus as shown in the diagram. Half fill the trough or bowl with water.
Fill the other measuring cylinder with water, and make sure that it stays filled with water when you turn it upside down.
When you are ready, add a 3 cm strip of magnesium ribbon to the flask, put the bung back into the flask as quickly as you can, and start the stopwatch.
Record the volume of hydrogen gas given off at suitable intervals (eg 10 seconds). Continue timing until no more gas appears to be given off.

Teaching notes

The equation for the reaction is: magnesium + hydrochloric acid → magnesium chloride + hydrogen

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Students follow the rate of reaction between magnesium and the acid, by measuring the amount of gas produced at 10 second intervals.

3 cm of magnesium ribbon typically has a mass of 0.04 g and yields 40 cm3 of hydrogen when reacted with excess acid. 50 cm3 of 1M hydrochloric acid is a six-fold excess of acid.

In this reaction, the magnesium and acid are gradually used up. However the acid is in excess, so it is mainly the loss of magnesium (surface area becomes smaller) that causes the change in the rate.

If a graph of volume (y-axis) against time (x-axis) is drawn, the slope of the graph is steepest at the beginning. This shows that the reaction is fastest at the start. As the magnesium is used up, the rate falls. This can be seen on the graph, as the slope becomes less steep and then levels out when the reaction has stopped (when no more gas is produced).

The reaction is exothermic, but the dilute acid is in excess and the rise in temperature is only of the order of 3.5˚C. There is some acceleration of the reaction rate due to the rise in temperature. Some students might notice the flask becoming slightly warm and they could be asked how this would affect the rate of reaction, and how they might adapt the experiment to make it a ‘fair test’.

With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below:

2 HCl (aq) + Zn (s) → ZnCl2 (aq) + H2 (g)

A sample of pure zinc with a mass of 5.98 g is reacted with excess hydrochloric acid and the (dry) hydrogen gas is collected at 25.0 ˚C and 742 mm Hg. What volume of hydrogen gas would be produced?

This is a “single state” problem, so we can solve it using the ideal gas law, PV = nRT. In order to find the volume of hydrogen gas (V), we need to know the number of moles of hydrogen that will be produced by the reaction. Our stoichiometry is simply one mole of hydrogen per mole of zinc, so we need to know the number of moles of zinc that are present in 5.98 grams of zinc metal. The temperature is given in centigrade, so we need to convert into Kelvin, and we also need to convert mm Hg into atm.

Conversions:

\[25.0\; C+273=298\; K \nonumber \]

\[(742\; mm\; Hg)\times \left ( \frac{1\; atm}{760\; mm\; Hg} \right )=0.976\; atm \nonumber \]

\[(5.98\; g\; Zn)\times \left ( \frac{1.00\; mol}{65.39\; g\; Zn} \right )=0.0915\; mol \nonumber \]

Substituting:

\[PV=nRT \nonumber \]

\[(0.976\; atm)\times V=(0.0915\; mol)(0.0821\; L\; atm\; mol^{-1}K^{-1})(298\; K) \nonumber \]

\[V=\frac{(0.0915\; mol)(0.0821\; L\; atm\; mol^{-1}K^{-1})(298\; K)}{(0.976\; atm)}=2.29\; L \nonumber \]

We can also use the fact that one mole of a gas occupies 22.414 L at STP in order to calculate the number of moles of a gas that is produced in a reaction. For example, the organic molecule ethane (CH3CH3) reacts with oxygen to give carbon dioxide and water according to the equation shown below:

2 CH3CH3 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (g)

An unknown mass of ethane is allowed to react with excess oxygen and the carbon dioxide produced is separated and collected. The carbon dioxide collected is found to occupy 11.23 L at STP; what mass of ethane was in the original sample?

Because the volume of carbon dioxide is measured at STP, the observed value can be converted directly into moles of carbon dioxide by dividing by 22.414 L mol–1. Once moles of carbon dioxide are known, the stoichiometry of the problem can be used to directly give moles of ethane (molar mass 30.07 g mol-1), which leads directly to the mass of ethane in the sample.

\[(11.23\; L\; CO_{2})\times \left ( \frac{1\; mol}{22.414\; L} \right )=0.501\; mol\; CO_{2} \nonumber \]

Reaction stoichiometry:

\[(0.501\; mol\; CO_{2})\times \left ( \frac{2\; mol\; CH_{3}CH_{3}}{4\; mol\; CO_{2}} \right )=0.250\; mol\; CH_{3}CH_{3} \nonumber \]

The ideal gas laws allow a quantitative analysis of whole spectrum of chemical reactions. When you are approaching these problems, remember to first decide on the class of the problem:

If it is a “single state” problem (a gas is produced at a single, given, set of conditions), then you want to use PV = nRT.
If it is a “two state” problem (a gas is changed from one set of conditions to another) you want to use \[\frac{P_{1}V_{1}}{n_{1}T_{1}}=\frac{P_{2}V_{2}}{n_{2}T_{2}} \nonumber \]
If the volume of gas is quoted at STP, you can quickly convert this volume into moles with by dividing by 22.414 L mol-1.

Once you have isolated your approach ideal gas law problems are no more complex that the stoichiometry problems we have addressed in earlier chapters.

An automobile air bag requires about 62 L of nitrogen gas in order to inflate. The nitrogen gas is produced by the decomposition of sodium azide, according to the equation shown below

2 NaN3 (s) → 2 Na (s) + 3 N2 (g)

What mass of sodium azide is necessary to produce the required volume of nitrogen at 25 ˚C and 1 atm?

When Fe2O3 is heated in the presence of carbon, CO2 gas is produced, according to the equation shown below. A sample of 96.9 grams of Fe2O3 is heated in the presence of excess carbon and the CO2 produced is collected and measured at 1 atm and 453 K. What volume of CO2 will be observed?

2 Fe2O3(s) + 3 C (s) → 4 Fe (s) + 3 CO2 (g)

The reaction of zinc and hydrochloric acid generates hydrogen gas, according to the equation shown below. An unknown quantity of zinc in a sample is observed
to produce 7.50 L of hydrogen gas at a temperature of 404 K and a pressure of 1.75 atm. How many moles of zinc were in the sample?

Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g)