50.0 g N204 (92.02 g/mol) react with 45.0 g N2H4 (32.05 g/mol) forming nitrogen gas, N2 (28.01 g/mol) and water, H20 (18.02 g/mol). What mass in grams of excess-reactant is
left over?

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.

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