Aluminum oxide produced : = 79.152 g
<h3>Further explanation</h3>
Given
46.5g of Al
165.37g of MnO
Required
Aluminum oxide produced
Solution
Reaction
2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)
mol = mass : Ar
mol = 46.5 : 27
mol = 1.722
mol = 165.37 : 71
mol = 2.329
mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776
MnO as a limiting reactant(smaller ratio)
So mol Al₂O₃ based on MnO as a limiting reactant
From equation , mol Al₂O₃ :
= 1/3 x mol MnO
= 1/3 x 2.329
= 0.776
Mass Al₂O₃ (MW=102 g/mol) :
= 0.776 x 102
= 79.152 g
Answer: A property closely related to an atom's mass number is its atomic mass. The atomic mass of a single atom is simply its total mass and is typically expressed in atomic mass units or amu. By definition, an atom of carbon with six neutrons, carbon-12, has an atomic mass of 12 amu.
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To find the mass you need to find the weight of a mol of the molecules by adding up the atomic mass.
N = 14.007 g/mol
H = 1.008 g/mol
S = 32.065 g/mol
O = 16 g/mol
2(14.007) + 8(1.008) + 32.065 + 4(16) = 132.143 g/mol
Now you know how much an entire mol weight you multiply it by how much you actually have
0.00456 * 132.143 = 0.603 g
Moles of electrons:
The moles of electrons that are transferred are 12F
A balanced equation:
2 moles of Aluminium metal react with excess copper(II) nitrate.
Given:
Moles of Aluminium = 2
As Aluminium goes from 0 to +3 oxidation state
And copper goes from +2 to 0
On balancing the number of electrons we get:
For 1 mole of Al 
is required.
Therefore for 2 moles of Al,
Total 
F mole of electrons
Where F= Faraday's constant= 96500 C
So, 12F moles of electrons are transferred.
Learn more about Faraday's Law here,
brainly.com/question/27985929
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