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
Answer:
a) After helping our partner, we should immediately report the incident to the lab manager or any person in charge of the emergencies occurring in the lab.
b) We should have a copy of the Material Safety Data Sheet to give to the responders. This is because the responder can identify what materials were being used by the person ans what other security measures need to be taken.
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
Answer:
Moles of NaCl formed is 6.0 moles
Explanation:
We are given the equation;
2 Na(s) + Cl₂(g) → 2 NaCl(s)
- Moles of Na is 6.0 moles
- Moles of Cl₂ is 4.0 moles
From the reaction;
2 moles of sodium reacts with 1 mole of chlorine gas to form 2 moles of NaCl
In this case;
6 moles of Na would require 3 moles of Cl₂, this means that chlorine gas is in excess.
Thus, the rate limiting reagent is sodium.
But, 2 moles of sodium reacts to form 2 moles of NaCl
Therefore;
Moles of NaCl = Moles of Na
= 6.0 moles
Thus, moles of sodium chloride produced is 6.0 moles
