You have the stoichiometric equation. This tells you unequivocally that an
18
⋅
g
mass of water, 1 mole, reacts with a
56.07
⋅
g
mass of quicklime to form a
74.09
⋅
g
mass of slaked lime.
If you don't from where I am getting these numbers, you should know, and someone will be willing to elaborate.
Here, you have formed
6.21
⋅
m
o
l
of quicklime which requires stoichiometric lime AND water. And thus you need a mass of
6.21
⋅
m
o
l
×
18.01
⋅
g
⋅
m
o
l
−
1
water
≅
88
⋅
g
.
In practice, of course I would not weigh out this mass. I would just pour
100
−
200
⋅
m
L
of water into the lime.
Answer:
Carbonates (CO3-2), phosphates (PO4-3) and sulfides (S-2) are insoluble.
The exceptions are the alkali metals and the ammonium ion.
<span>This is rather a case of purification of impure copper or extraction of Cu from its alloy. You need to place the impure copper rod on the positive electrode (Anode) usualyy made of carbon rod, whereby oxidation reaction takes place: Cu (s) -------> Cu2+ (aq) + 2e- That is to say the impure cooper dissolves into solution. The copper (II) ions move to the negative electrode (cathode) usually made of pure copper rod. At the cathode, the Cu2+ ions are reduced : Cu2 (aq)+ + 2e- -----> Cu(s). That is to say the copper (II) ions are deposited as solid copper atoms onto the cu-rod electrode. In this way impure copper is deposited as pure copper onto the copper cathode</span>
