calculate expected yield of albr3 from 50g of aluminium when reacted with excess of bromine
Answers
Answer:
From your stoichiometrically balanced equation you note the 1:1 equivalence between moles of metal, and moles of salt: one mole of aluminum yields one mole of alumium tribromide given 100% yield.
Moles of aluminum
=
6.0
⋅
g
26.98
⋅
g
⋅
m
o
l
−
1
=
0.222
⋅
m
o
l
.
Had all the metal reacted, there would be
0.222
⋅
m
o
l
×
266.29
⋅
g
⋅
m
o
l
−
1
=
59.12
⋅
g
A
l
B
r
3
Thus yield
=
Recovered mass
Theoretical mass
=85
Answer:
Explanation:
If I understand your question correctly, the answer will be
197 g
.
The trick here is to realize that the combined mass of the reactants must be equal to the combined mass of the products
→
think the Law of conservation of mass here.
You know that you're mixing
20 g
of aluminium metal and
200 g
of bromine, but that the reaction only consumes
what is available
200 g
−
what is left behind
23 g
=
177 g
of bromine. This means that the combined mass of the reactants will be
20 g + 177 g = 197 g
Since the reaction will only produce aluminium bromide, you can say that the mass of the product must be equal to
197 g
.
CHECK THE RESULT USING MOLES
If you want, you can check the result by using the balanced chemical equation
2
Al
(
s
)
+
3
Br
2
(
l
)
→
Al
2
Br
6
(
a
q
)
Use the molar masses of the two reactants to calculate how many moles of each take part in the reaction--keep in mind that only
177 g
of bromine react!
20
g
⋅
1 mole Al
27.0
g
=
0.74 moles Al
177
g
⋅
1 mole Br
2
159.8
g
=
1.11 moles Br
2
Aluminium reacts with bromine in a
2
:
3
mole ratio
0.74
moles Al
⋅
3 moles Br
2
2
moles Al
=
1.11 moles Al
and produces aluminium bromide in a
2
:
1
mole ratio
0.74
moles Al
⋅
1 mole Al
2
Br
6
2
moles Al
=
0.37 moles Al
2
Br
6
Finally, to convert this to grams, use the molar mass of aluminium bromide
0.37
moles Al
2
Br
6
⋅
533.4 g
1
mole Al
2
Br
6
≈
197 g