An Introduction to Chemical Science Part 4

15). At the other end have a philosopher's lamp-tube.Observing the usual precautions, light the gas and hold over it a receiver, till quite a quant.i.ty of moisture collects. All water was taken from the gas by the dryer, CaCl2. What is, therefore, the product of burning H in air? Complete this equation and explain it: 2H + O = ? Figure 16 shows a drying apparatus arranged to hold CaCl2.

[Fig. 15][Fig. 16]

37. Explosiveness of H.

Experiment 25. -- Fill a soda-water bottle of thick gla.s.s with water, invert it in a pneumatic trough, and collect not over 1/4 full of H. Now remove the bottle, still inverted, letting air in to fill the other 3/4. Mix the air and H by covering the mouth of the bottle with the hand, and shaking well; then hold the mouth of the bottle, slightly inclined, in a flame. Explain the explosion which follows. If 3/4 was air, what part was O? What use did the N serve? Note any danger in exploding H mixed with pure O. What proportions of O and H by volume would be most dangerously explosive? What proportion by weight?

By the rapid union of the two elements, the high temperature suddenly expanded the gaseous product, which immediately contracted; both expansion and contraction produced the noise of explosion.

38. Pure H Is a Gas without Color, Odor, or Taste.

--It is the lightest of the elements, 14 1/2 times as light asair. It occurs uncombined in coal-mines, and some other places, but the readiness with which it unites with other elements, particularly O, prevents its acc.u.mulation in large quant.i.ties. It const.i.tutes two-thirds of the volume of the gases resulting from the decomposition of water, and one-ninth of the weight. Compute the latter from its symbol. It is a const.i.tuent of plants and animals, and some rocks. Considering the volume of the ocean, the total amount of H is large. It can be separated from H2O by electrolysis, or by C, as in the manufacture of water gas.

When burned with O it forms H2O. Pure O and H when burning give great heat, but little light. The oxy-hydrogen blow-pipe (Fig.

17) is a device for producing the highest temperatures of combustion. It has O in the inner tube and H in the outer. Why would it not be better the other way? These unite at the end, and are burned, giving great heat. A piece of lime put into the flame gives the brilliant Drummond or calcium light.

Chapter IX. UNION BY WEIGHT.

39. In the Equation --

Zn + 2 HCl = ZnCl2 + 2 H 65 + 73 = 136 + 2

65 parts by weight of Zn are required to liberate 2 parts by weight of H; or, by using 65 g Zn with 73 g HCl, we obtain 2 g H. If twice as much Zn (130 g) were used, 4 g H could be obtained, with, of course, twice as much HCl. With 260 g. Zn, how much H could be liberated?

A proportion may be made as follows:--

Zn given : Zn required :: H given : H required.

65 : 260 :: 2 : x.

[footnote: Given, as here used, means the weight called for by the equation; required means that called for by the question.]

Solving, we have 8 g H.

How much H is obtainable by using 5 g Zn, as in the experiment?

To avoid error in solving similar problems, the best plan is as follows:--

Zn + 2HCl = ZnCl2 + 2 H | 65:5::2:x 65 2 | 65 x = 10 5 x | x = 10/65 = 2/13Ans. 2/13 g.

The equation should first be written; next, the atomic or molecular weights which you wish to use, and only those, to avoid confusion; then, on the third line, the quant.i.ty of the substance to be used, with underneath the substance wanted. The example above will best how this. This plan will prevent the possibility of error. The proportion will then be:--

a given : a required :: b given : b required.

How much Zn is required to produce 30 g. H?

Zn + 2HCl = ZnCl2 + 2H | 2:30::65:x 65 2 | 2x = 1950 x 30 | x = 975 Ans. 975 g. Zn.

Solve:--

(1) How much Zn is necessary for 14 g. H?

(2) How many pounds of Zn are necessary for 3 pounds of H?

(3) How many grams of H from 17 g. of Zn?

(4) How many tons of H from 1/2 ton of Zn?

Suppose we wish to find how much chlorhydric acid--pure gas-- will give 12 g. H. The question involves only HCl and H. Arrange as follows:--

Zn + 2HCl = ZnCl2 + 2 H | H giv. : H req. :: HCl giv. : HCl req.

73 2 | 2 : 12 :: 73 x x 12 | 2x=876 x=438 Ans. 438 g. HCl.

Solve:--

(1) How much HCl is needed to produce 100 g. H?

(2) How much H in 10 g. HCl?

(3) How much ZnCl2 is formed by using 50 g. HCl? The question is now between HCl and ZnCl2.

Zn + 2HC1 = ZnCl2 + 2H 73 136 | Arrange the proportion, and solve.

50 x

Suppose we have generated H by using H2S04: the equation is Zn + H2S04 = ZnSO4 + 2 H. There is the same relation as before between the quant.i.ties of Zn and of H, but the H2S04 and ZnS04 are different.

How much H2SO4 is needed to generate 12 g. H?

Zn + H2SO4 = ZnS04 + 2 H 98 2 | Make the proportion, and solve x 12

Solve:--

(1) How much H in 200 g. H2S04?

(2) How much ZnS04 is produced from 200 g. H2S04?

(3) How much H2S04 is needed for 7 1/2 g H?

(4) How much Zn will 40 g. H2SO4 combine with?

(5) How much Fe will 40 g. H2SO4 combine with?

(6) How much H can be obtained by using 75 g Fe?

These principles apply to all reactions. Suppose, for example, we wish to get l0 g. of O: how much KClO3 will it be necessary to use?

The reaction is:--

KClO3 = KCl + O3 | 48 : 10 :: 122.5 : x 122.5 48 | x 10 | Ans. 25.5+ g. KClO3.

The pupil should be required to make up problems of his own, using various reactions, and to solve them.

CHAPTER X.

CARBON.