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.