Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles =
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O =
= 0.13moles
Number of moles of NH₃ =
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
To count the number of valence electrons we look at the electronic configuration and add the electrons form the electronic shell with the highest principal quantum number.
Rb: [Kr] 5s¹ - 1 valence electron
Xe: [Kr] 5s² 4d¹⁰ 5p⁶ - 8 valence electrons
Sb: [Kr] 5s² 4d¹⁰ 5p³ - 5 valence electrons
I: [Kr] 5s² 4d¹⁰ 5p⁵ - 7 valence electrons
In: [Kr] 5s² 4d¹⁰ 5p¹ - 3 valence electrons
Rank from most to fewest valence electrons:
Xe > I > Sb > In > Rb
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The number of grams is 10