Answer:
A. The limiting reagent is CuO
B. 5.31g of N2 is produced.
C. Percentage yield is 86.8%
D. The mass of the excess reactant (i.e NH3) that remain is 2.61g
Explanation:
A. Step 1:
The balanced equation for the reaction.
2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(l)
A. Step 2:
Determination of the masses of NH3 and CuO that reacted from the balanced equation. This is illustrated below:
2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(l)
Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34g
Molar Mass of CuO = 63.5 + 16 = 79.5g/mol
Mass of CuO from the balanced equation = 3 x 79.5 = 238.5g
From the balanced equation above,
34g of NH3 reacted.
238.5g of CuO reacted.
A. Step 3:
Determination of the limiting reagent.
Now, let us consider using all the 9.05 g of NH3 given to see if there will be leftover for CuO.
From the balanced equation above,
34g of NH3 reacted with 238.5g of CuO.
Therefore, 9.05 g of NH3 will react with = ( 9.05x238.5)/34 = 63.48g of CuO.
The mass of CuO obtained is far higher than 45.2 g given from the question. Therefore it not acceptable.
Now let consider using all 45.2 g of CuO to see if there will be leftover for NH3.
From the balanced equation above,
34g of NH3 reacted with 238.5g of CuO.
Therefore, Xg of NH3 will react 45.2 g of CuO i.e
Xg of NH3 = (34x45.2)/238.5
Xg of NH3 = 6.44g
Now, we can see that there are leftover for NH3 as only 6.44g of NH3 reacted out of 9.05g given.
Therefore, the limiting reagent is CuO
B. Step 1:
Determination of the mass of CuO that reacted and the mass of N2 produced from the balanced equation. This is illustrated below:
2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(l)
Molar Mass of CuO = 63.5 + 16 = 79.5g/mol
Mass of CuO from the balanced equation = 3 x 79.5 = 238.5g
Molar Mass of N2 = 2x14 = 28g/mol
From the balanced equation,
238.5g of CuO reacted.
28g of N2 is produced.
B. Step 2.
Determination of the mass of N2 produce when 9.05 g of NH3 reacted with 45.2 g of CuO. This is illustrated below:
From the balanced equation above,
238.5g of CuO reacted to produce 28g of N2.
Therefore, 45.2g of CuO will react to produce = (45.2 x 28)/238.5 = 5.31g of N2.
C. Determination of the percentage yield of N2. This is shown below:
Actual yield of N2 = 4.61 g
Theoretical yield = 5.31g
Percentage yield =?
Percentage yield = Actual yield /Theoretical yield x100
Percentage yield = 4.61/5.31 x 100
Percentage yield = 86.8%
D. Determination of the mass of the excess reactant that remains.
From the calculations made above, the excess reactant is NH3.
The mass of NH3 that remain after the reaction is given by:
Mass of NH3 given = 9.05 g
Mass of NH3 that reacted = 6.44g
Mass of NH3 that remains = (Mass of NH3 given) - (Mass of NH3 that reacted)
Mass of NH3 that remains = 9.05 - 6.44
Mass of NH3 that remains = 2.61g