Respuesta :
Approx.
15
⋅
g
of metal.
Explanation:
Moles of iron oxide,
F
e
2
O
3
=
21.6
⋅
g
159.69
⋅
g
⋅
m
o
l
=
0.135
⋅
m
o
l
with respect to the oxide.
But by the composition of the oxide, there are thus
2
×
0.135
⋅
m
o
l
×
55.8
⋅
g
⋅
m
o
l
−
1
iron metal
=
?
?
g
.
An extraordinary percentage of our budgets goes into rust prevention;
iron(III) oxide
is only part of the redox chemistry. Once you put up a bridge or a skyscraper, which is inevitably steel-based structure, it will begin to corrode.
15
⋅
g
of metal.
Explanation:
Moles of iron oxide,
F
e
2
O
3
=
21.6
⋅
g
159.69
⋅
g
⋅
m
o
l
=
0.135
⋅
m
o
l
with respect to the oxide.
But by the composition of the oxide, there are thus
2
×
0.135
⋅
m
o
l
×
55.8
⋅
g
⋅
m
o
l
−
1
iron metal
=
?
?
g
.
An extraordinary percentage of our budgets goes into rust prevention;
iron(III) oxide
is only part of the redox chemistry. Once you put up a bridge or a skyscraper, which is inevitably steel-based structure, it will begin to corrode.
Answer:
The amount of chlorine present in iron (iii) chloride is 656.59 g
Explanation:
Given amount of iron (iii) chloride = 1000 g
Molar mass of Iron (iii) chloride = 162.2 g/mol
Number of moles of Iron (iii) chloride = [tex]\displaystyle \frac{1000}{162.2}\textrm{ mole} = 6.16 \textrm{ mole}[/tex]
Number of moles of chlorine present = (3 [tex]\times[/tex] 6.16 ) mole = 18.49 mole
Atomic mass of chlorine = 35.5 g/mol
Amount of chlorine present = 18.49 mole [tex]\times[/tex] 35.5 g/mol = 656.59 g
Amount of chlorine present = 656.59 g