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.

where are the choice options?

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.