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
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