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
I think the awnser to your question is C
Weight percentage of nitrogen can be calculated using the following rule:
weight percentage of nitrogen = (weight of nitrogen / weight of urea) x 100
From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of nitrogen = 14 grams
molecular mass of hydrogen = 1 grams
molecular mass of oxygen = 16 grams
therefore:
mass of nitrogen in urea = 2(14) = 28 grams
mass of urea = 12 + 2(14) + 4(1) + 16 = 60 grams
Substitute with the masses in the equation to get the percentage:
weight percentage of nitrogen = (28/60) x 100 = 46.667%
Option B. <span>Rb2O + Cu(C2H3O2)2 → 2RbC2H3O2 + CuO is the correct answer. HOPE IT HELPS</span>
It may just be my internet but the picture is just white. maybe try reposting this..?