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<u>Answer:</u> The molecules of oxygen gas that will be reduced to water are 42 molecules
<u>Explanation:</u>
We are given:
The substance having highest positive potential will always get reduced and will undergo reduction reaction. Here, oxygen will undergo reduction reaction will get reduced.
will undergo oxidation reaction and will get oxidized.
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
The half reactions follows:
<u>Oxidation half reaction:</u> ( × 4)
<u>Reduction half reaction:</u> ( × 6)
<u>Overall reaction:</u>
We are given:
Molecules of = 28
By Stoichiometry of the reaction:
4 molecules of reacts with 6 molecules of oxygen gas
So, 28 molecules of will react with =
molecules of oxygen gas
Hence, the molecules of oxygen gas that will be reduced to water are 42 molecules
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