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.Start with the number of grams of each element, given in the problem.
2.Convert the mass of each element to moles using the molar mass from the periodic table.
3.Divide each mole value by the smallest number of moles calculated.
4.Round to the nearest whole number. This is the mole ratio of the elements and is.
Answer:
Explanation:
Hello there!
In this case, since the reaction (A->B) have an initial amount of pure 4-aminobenzoic acid, the first step to compute the theoretical yield is to solve the following stoichiometric setup:
Whereas A stands for 4-aminobenzoic acid and B for the benzocaine. Moreover, we compute the percent yield by dividing the actual yield (0.318 g) by the theoretical one (0.365 g):
Best regards!
Answer is: <span>- delta G.
</span>The change in Gibbs free energy (ΔG), at constant temperature and pressure, is: <span>ΔG=ΔH−TΔS.
</span>ΔH<span> is the change in enthalpy.
</span>ΔS is change in entropy.
T is temperature of the system.
When ΔG is negative, a reaction (<span>occurs without the addition of external energy)</span><span> will be spontaneous (</span>exergonic).
Number 2 is the mass of an electron
