During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that the following side reaction (ii) occurs. (i) N2O4(l) 2 N2H4(l) 3 N2(g) 4 H2O(g) (ii) 2 N2O4(l) N2H4(l) 6 NO(g) 2 H2O(g) In one experiment 12.7 g of NO formed when 101.1 g of each reactant was used. What is the highest percent yield of N2 that can be expected

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

N₂O₄ is the limiting reactant

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%

your answer is c) plasma

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