Answer:
Maximum expected yield = 87.2%
Explanation:
Equations of reactions:
Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)
Side reaction: 2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)
Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol
In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.
101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄
<em>N₂O₄ is the limiting reactant</em>
101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ = 46.15 g of N₂
In the side reaction, (6 * 30 g) of NO is produced from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄
12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce
mass of N₂O₄ used = 12.98 g
mass of N₂H₄ used = 2.26 g
mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g
mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g
In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂
88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄
N₂O₄ is the limiting reactant.
88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ = 40.23 g of N₂
Percentage yield = (theoretical yield/actual yield) * 100%
Percentage yield = (40.23/46.15) * 100% = 87.2%
Therefore, maximum expected yield = 87.2%
