the Equ

nCl2 + 2 H 65

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

:: H given : H requir

weight called for by the equation; requi

, we ha

ble by using 5 g Zn,

g similar problems, the

5::2:x 65 2 | 65 x = 10 5 x

avoid confusion; then, on the third line, the quantity of the substance to be used, with underneath the substance want

uired :: b give

required to p

30::65:x 65 2 | 2x = 1950 x

lv

Zn is necessa

of Zn are necessar

grams of H fr

ons of H from

acid-pure gas- will give 12 g. H. The questio

H | H giv. : H req.

2 : 12

2x=87

438 g

lv

l is needed to

uch H in

y using 50 g. HCl? The questio

73 136 | Arrange the pr

2S04 = ZnSO4 + 2 H. There is the same relation as before between t

is needed to ge

O4 = ZnS

the proporti

lv

ch H in 20

for 7 1/2 g H? (4) How much Zn will 40 g. H2SO4 combine with? (5) How much Fe

for example, we wish to get l0 g. of O: how much K

:: 122.5 : x 122.5 48 | x

ke up problems of his own, using v