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Sagot :
Answer:
0.347
Explanation:
Al is the limiting reagent.
Explanation:
The stoichiometry of the reaction is such that 4 moles of Al are required for every 3 moles of diatomic oxygen. This means that if the ratio of Al to diatomic oxygen is greater than 4/3, then the oxygen is the limiting reagent. If the ratio of Al to diatomic oxygen is less than 4/3, then Al is the limiting reagent.
The problem state that the ratio of Al to diatomic oxygen is
0.32
0.26
=
16
13
=
48
39
<
52
39
=
4
3
The ration of Al to diatomic oxygen is less than 4/3, so Al is the limiting reagent.
We can also see this if we ask the question "How much Aluminum" is required to completely react 0.26 moles of diatomic oxygen??
0.26
moles
O
2
×
4
moles
A
l
3
moles
O
2
≈
0.347
moles
A
l
So it would take 0.347 moles of Al to completely react with all of the oxygen, however there is only 0.32 moles of Aluminum present, so there is not enough Al to react with all of the oxygen, and so we say that the Al is the limiting reagent.
The mass of Aluminum are needed to produce 21.6 grams of Aluminum oxide (Al₂O₃) is
What is mass?
Mass of any element is equal to the product of its number of moles and molar mass.
For the compound, Aluminum oxide (Al₂O₃)
The atomic masses of Aluminum is 23 and oxygen is 16.
Molecular or molar mass = ( 2 × 23 ) + ( 3 × 16 ) units
M = 46 + 48 = 94 units
The mass of Aluminum in 1 mole of aluminum oxide ,46 grams in 94 grams
Mass of Aluminum in the given 21.6 g of Aluminum oxide will be;
94 grams of Aluminum oxide contains 46 grams of Aluminum.
21.6 grams of Aluminum oxide will contain ( 46 / 94 ) × 21.6 = 10.57g of Aluminum.
Therefore, 10.57g of Aluminum are needed to produce 21.6 grams of Aluminum oxide (Al₂O₃)
Learn more about mass.
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