Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N2O4 and 45.0 g N2H4. Some p

Answer:

N2O4 is identified as the limiting reactant.

We can expect to produce 45.6 grams of N2.

Explanation:

Step 1: Given data

Mass of N2O4 = 50.0 grams

Mass of N2H4 = 45.0 grams

Molar mass of N2O4 = 92.02 g/mol

Molar mass of N2H4 = 32.05 g/mol

Step 2: The balanced reaction

N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)

Step 3: Calculating moles

Moles = mass / molar mass

Moles N2O4 = 50.0 grams / 92.02 g/mol

Moles N2O4 = 0.543 moles

Moles N2H4 = 45.0 grams / 32.05 g/mol

Moles N2H4 = 1.404 moles

Step 4: Determine the limiting reactant

For each mole of N2O4, two moles of N2H4 are required to generate three moles of N2 and four moles of H2O.

N2O4 will completely react (0.543 moles), while N2H4 is in excess. A total of 1.086 moles of N2H4 will be consumed, leaving 0.318 moles of N2H4 remaining.

Step 5: Calculating the moles of nitrogen (N2)

For 0.543 moles of N2O4, we derive 3 * 0.543 = 1.629 moles of N2.

Step 6: Calculating the mass of N2

Mass of N2 = moles of N2 * molar mass of N2

Mass of N2 = 1.629 moles * 28.0 g/mol

Mass of N2 = 45.6 grams of N2.

Thus, we will yield 45.6 grams of N2.

Answer:

N2O4 serves as the limiting reactant.

B. The mass of nitrogen produced is 45.67 g.

Explanation:

The reaction is represented by the following balanced equation:

N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)

A. To identify the limiting reactant, we must first calculate the amounts of N2O4 and N2H4 utilized from the balanced equation as shown:

Molar mass of N2O4 = 92.02 g/mol

Molar mass of N2H4 = 32.05 g/mol

Amount of N2H4 reacting from the equation = 2 x 32.05 = 64.1 g

Now, assuming we use the entire 50g sample of N2O4, we can check how much N2H4 is necessary.

According to the balanced equation,

92.02g of N2O4 requires 64.1g of N2H4.

Thus, 50g of N2O4 will need (50 x 64.1)/92.02 = 34.83g of N2H4.

Comparing this usage of N2H4 (34.83g) to the available mass (45g) shows that N2H4 is in excess, with a leftover of (45 - 34.83 = 10.17g) N2H4.

Thus, N2O4 is the limiting reactant.

B. To find the nitrogen mass produced, the limiting reactant will guide us.

The reaction is represented as:

N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)

Molar mass of N2 = 2 x 14.01 = 28.02 g/mol

Mass of N2 according to the balanced equation = 3 x 28.02 = 84.06 g

Knowing that 92.02g of N2O4 yields 84.06g of N2, we can calculate for 50g of N2O4:

(50 x 84.06) / 92.02 = 45.67g of N2.

Thus, from the reaction between 50.0g of N2O4 and 45.0g of N2H4, 45.67g of nitrogen will be generated.