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
1) This is a definition.
2) Protons are given by the bottom number (since atomic number = number of protons).
3) Neutrons = (mass number)-(atomic number), which are the top and bottom numbers, respectively.
4) Nuclear fusion involves combining two things together, which is only reflected by the last option.
5) This is a fact.
6) This is a fact.
7) This is a fact.
8) This is a fact.
9) The correct option is the explanation.
The molar mass of CO2 can be calculated as follows;
CO2 — 12 + (16x2) = 12+ 32 = 44 g
Therefore molar mass of CO2 is 44 g/mol
In 44 g of CO2 there’s 1 mol of CO2
Then 1 g of CO2 there’s 1/44 mol of CO2
Therefore in 78.3 g of CO2 there’s — 1/44 x 78.3 =1.78 mol of CO2