this is liquid water in the form of droplets that have condoned from atmospheric water vapor and than precipitated
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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
The mentioned molecule with formula, CH₃NO where no bond is found between N and O can be depicted as formamide is shown in the attachment below.
The formal charges = Number of valence electrons for the atom (V) - the number of electrons in lone pairs (N) - 1/2 (number of electrons in bond pairs, B)
FC = V - N - B/2
Thus, there is a need to calculate valence electrons, electrons in lone pairs, and the number of electrons in bond pairs for each atom in the mentioned molecule.
V or valence electrons on C = 4e, on H = 1e, on N = 5e, and on O = 6e.
N or electrons in lone pairs on C = 0e, on H = 0e, on N = 2e, and on O = 4e.
B or number of electrons in bond pairs for C = 8e, for H = 2e, for N = 6e, and for O = 4e.
Thus, the formal charges for each will be,
C = 4 - 0 - (8/2) = 0
H = 1 - 0 - (2/2) = 0
N = 5 - 2 - (6/2) = 0
O = 6 - 4 - (4/2) = 0
Lewis dot structure for the given molecule is given in the attachment below:
