Answer:
Option A. LR = N2O4, 45.7g N2 formed
Step-by-step explanation:
The balanced equation for the reaction is given below:
N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)
Next, we shall determine the masses of N2O4 and N2H4 that reacted and mass of N2 produced from the balanced equation. This is illustrated below:
Molar mass of N2O4 = 92.02 g/mol
Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02 g
Molar mass of N2H4 = 32.05 g/mol
Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g
Molar mass of N2 = 2x14.01 = 28.02g/mol
Mass of N2 from the balanced equation = 3 x 28.02 = 84.06g
Summary:
From the balanced equation above,
92.02g of N2O4 reacted with 64.1g of N2H4 to produce 84.06g of N2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
92.02g of N2O4 reacted with 64.1g of N2H4.
Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.
From the calculations made above, we can see that only 34.83g out 45g of N2H4 is required to react completely with 50g of N2O4.
Therefore, N2O4 is the limiting reactant and N2H4 is the excess reactant.
Finally, we shall determine the mass of N2 produced from the reaction.
In this case the limiting reactant will be used as it will produce the maximum yield of N2 since all of it is used up in the reaction.
The limiting reactant is N2O4 and the mass N2 produced can be obtained as illustrated below:
From the balanced equation above,
92.02g of N2O4 reacted to produce 84.06g of N2.
Therefore 50g of N2O4 will react to produce = (50 x 84.06)/92.02 = 45.7g of N2.
Therefore, 45.7g of N2 were produced from the reaction.
At the end of the day,
The limiting reactant is N2O4 and 45.7g of N2 were produced from the reaction.