Answer:
The excess reactant is N2H4 and the leftover mass is 10.17g.
Explanation:
Step 1:
The balanced equation for the reaction.
N2O4 + 2N2H4 —> 3N2 + 4H2O
Step 2
Determination of the masses of N2O4 and N2H4 that reacted from the balanced equation:
Molar mass of N2O4 = 92.02 g/mol
Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02g
Molar mass of N2H4 = 32.05 g/mol
Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g
From the balanced equation above, 92.02g of N2O4 reacted with 64.1g of N2H4.
Step 3:
Determination of the excess reactant. This is illustrated below:
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 of N2H4 reacted out of 45g that was given. Therefore, N2H4 is the excess reactant.
Step 4:
Determination of the mass of excess reactant that is leftover.
The excess reactant is N2H4 and the leftover mass can be obtained as follow:
Mass of N2H4 given = 45g
Mass of N2H4 that reacted = 34.83g
Leftover mass of N2H4 =..?
Leftover mass of N2H4 = (Mass of N2H4 given) – (Mass of N2H4 that reacted)
Leftover mass of N2H4 = 45 – 34.83
Leftover mass of N2H4 = = 10.17g.
