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 = 2/13 Ans. 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 quantity 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 quantities 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.