Question

Determine the percent yield for the reaction between 15.0 g of N2 and 15.0 g of H2 if 10.5 g of NH3 is produced.

Answer 1

To determine the percent yield for a reaction, compare the actual yield to the theoretical yield. Use the balanced equation and molar masses to convert masses to moles. Calculate the stoichiometric ratio and use it to find the moles of NH3 that should be produced. Finally, calculate the percent yield using the formula.

Step-by-step explanation:

To determine the percent yield for a reaction, you need to compare the actual yield to the theoretical yield. Theoretical yield is the maximum amount of product that can be formed based on the stoichiometry of the balanced equation. Actual yield is the amount of product obtained from the reaction.

In this case, the balanced equation is N2 + 3 H2 → 2 NH3. From the equation, we can calculate the molar masses of N2, H2, and NH3. Using the given values:

• Mass of N2 = 15.0 g
• Mass of H2 = 15.0 g
• Mass of NH3 produced = 10.5 g

Step 1: Convert the masses of N2, H2, and NH3 to moles using their molar masses.

Step 2: Determine the stoichiometric ratio of N2 to NH3. In the balanced equation, the ratio is 1:2. Use this ratio to calculate the moles of NH3 that should be produced if all the N2 reacted.

Step 3: Calculate the percent yield using the formula:

Percent Yield = (Actual Yield / Theoretical Yield) × 100

Substitute the values into the formula to find the percent yield.

Answer 2

Step-by-step explanation:

Nitrogen will react with hydrogen to give ammonia. The balanced reaction between them is:

N₂ + 3 H₂ ----> 2 NH₃

In our problem, 15.0 g of N₂ are reacting with 15.0 g of H₂. Before we find the percent yield we will have to determine the limiting reactant.

First, let's find the number of moles of NH₃ that will be produced by 15.0 g of N₂ (with excess H₂).

molar mass of N₂ = 2 * 14.01 g/mol

molar mass of N₂ = 28.02 g/mol

mass of N₂ = 15.0 g

moles of N₂ = 15.0 g/(28.02 g/mol)

moles of N₂ = 0.535 moles

N₂ + 3 H₂ ----> 2 NH₃

According to the coefficients of the reaction, 1 mol of N₂ will produce 2 moles of NH₃. The molar ratio between N₂ and NH₃ is 1 to 2. We can use that ratio to find the number of moles of NH₃ that will be produced by 0.535 moles of N₂ (or 15.0 g of it).

1 mol of N₂ = 2 moles of NH₃

moles of NH₃ = 0.535 moles of N₂ * 2 moles of NH₃/(1 mol of N₂)

moles of NH₃ = 1.07 moles

So we found that 15.0 g of N₂ will yield 1.07 moles of NH₃. Let's do something similar to find the number of moles of NH₃ that will be produced by 15.0 g of H₂.

molar mass of H₂ = 2 * 1.01 g/mol

molar mass of H₂ = 2.02 g/mol

mass of H₂ = 15.0 g

moles of H₂ = 15.0 g/(2.02 g/mol)

moles of H₂ = 7.43 moles

3 moles of H₂ = 2 moles of NH₃

moles of NH₃ = 7.43 moles of H₂ * 2 moles of NH₃/(3 moles of H₂)

moles of NH₃ = 4.95 moles

So, 15.0 g of H₂ (reacting with excess N₂) will produce 4.95 moles of NH₃. And 15.0 g of N₂ will yield 1.07 moles of NH₃. So the reactant that is limiting our reaction is N₂ and H₂ is in excess.

N₂ = limiting reactant H₂ = excess

Now that we know the number of moles of NH₃ we can convert them into grams using its molar mass and finally find the percent yield of the reaction.

molar mass of NH₃ = 1 * 14.01 g/mol + 3 * 1.01 g/mol

molar mass of NH₃ = 17.04 g/mol

mass of NH₃ = 1.07 moles * 17.04 g/mol

mass of NH₃ = 18.2 g = theoretical yield

actual yield of NH₃ = 10.5 g

% yield = actual yield/theoretical yield * 100

% yield = 10.5 g/18.2 g * 100

% yield = 57.7 %

Answer: the percent yield for the reaction is 57.7 %.