Problem 147 - Introductory Chemistry Atoms First

Question

For the reaction \(\mathrm{NH}_{3}+\mathrm{NO} \rightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}:\) (a) Balance the equation. (b) If you react \(15.0 \mathrm{~g} \mathrm{NH}_{3}\) with \(22.0 \mathrm{~g} \mathrm{NO}\) and you produce \(13.3 \mathrm{~g} \mathrm{~N}_{2}\), what is your percent yield?

Step-by-Step Solution

Answer

The balanced equation for the given reaction is \(2\mathrm{NH}_{3}+\mathrm{NO}\rightarrow \mathrm{N}_{2}+3\mathrm{H}_{2}\mathrm{O}\). For the provided amounts of reactants, with \(\mathrm{NH}_{3}\) as the limiting reactant, the theoretical yield of \(\mathrm{N}_{2}\) is 12.35 g. Based on the given actual yield of 13.3 g, the calculated percent yield is 107.7%, which suggests an error in the actual yield provided.

Step 1: 1. Balance the equation

To balance the equation, we need to adjust the coefficients of the reactants and products, so that the number of atoms of each element is equal on both sides of the reaction. The given reaction is: \(\mathrm{NH}_{3} + \mathrm{NO} \rightarrow \mathrm{N}_{2} + \mathrm{H}_{2}\mathrm{O}\) There are two nitrogen atoms on the right side while there is only one nitrogen atom on each of the reactants. Changing the coefficient of \(\mathrm{NH}_{3}\) to 2 will balance the nitrogen atoms: \[\underline{\bold{2}}\mathrm{NH}_{3}+\mathrm{NO}\rightarrow \mathrm{N}_{2}+\mathrm{H}_{2}\mathrm{O}\] Now, amend the coefficients of water to balance the hydrogen atoms: \(2\mathrm{NH}_{3}+\mathrm{NO}\rightarrow \mathrm{N}_{2}+\underline{\bold{3}}\mathrm{H}_{2}\mathrm{O}\) The balanced equation is: \(2\mathrm{NH}_{3}+\mathrm{NO}\rightarrow \mathrm{N}_{2}+3\mathrm{H}_{2}\mathrm{O}\)

Step 2: 2. Determine the limiting reactant

Given amounts of reactants: - 15.0 g of \(\mathrm{NH}_{3}\) - 22.0 g of \(\mathrm{NO}\) First, calculate the moles of each reactant. Molar mass of \(\mathrm{NH}_{3} = 14.01 (\mathrm{N}) + 3 \times 1.01(\mathrm{H}) = 17.03 \text{ g/mol}\) Moles of \(\mathrm{NH}_{3} = \frac{15.0 \text{ g}}{17.03 \text{ g/mol}} = 0.881 \text{ mol}\) Molar mass of \(\mathrm{NO} = 14.01 (\mathrm{N}) + 16.00 (\mathrm{O}) = 30.01 \text{ g/mol}\) Moles of \(\mathrm{NO} = \frac{22.0 \text{ g}}{30.01 \text{ g/mol}} = 0.733 \text{ mol}\) Now, divide the moles of each reactant by their respective coefficients in the balanced equation and find the lowest ratio: \(\frac{\text{moles of NH}_{3}}{2} = \frac{0.881}{2} = 0.441\) \(\frac{\text{moles of NO}}{1} = \frac{0.733}{1} = 0.733\) As 0.441 is the lowest ratio, \(\mathrm{NH}_{3}\) is the limiting reactant.

Step 3: 3. Calculate the theoretical yield

Using the stoichiometry of the balanced equation, calculate the theoretical amount of \(\mathrm{N}_{2}\) that can be produced: \(2\mathrm{NH}_{3}+\mathrm{NO}\rightarrow \mathrm{N}_{2}+3\mathrm{H}_{2}\mathrm{O}\) From the balanced equation, two moles of \(\mathrm{NH}_{3}\) give one mole of \(\mathrm{N}_{2}\). So, we can calculate the theoretical moles of \(\mathrm{N}_{2}\): Theoretical moles of \(\mathrm{N}_{2} = \frac{1}{2} \times \text{moles of NH}_{3}\) Theoretical moles of \(\mathrm{N}_{2} = \frac{1}{2} \times 0.881 = 0.441 \text { mol}\) Now, convert moles of \(\mathrm{N}_{2}\) to mass using its molar mass: Molar mass of \(\mathrm{N}_{2} = 2 \times 14.01 = 28.02 \text{ g/mol}\) Theoretical yield of \(\mathrm{N}_{2} = 0.441 \text{ mol} \times 28.02 \text{ g/mol} = 12.35 \text{ g}\)

Step 4: 4. Calculate the percent yield

We are given that the actual yield of \(\mathrm{N}_{2}\) is 13.3 g. To calculate the percent yield, use the formula: Percent yield = \(\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\) Percent yield = \(\frac{13.3 \text{ g}}{12.35 \text{ g}} \times 100\% = 107.7\%\) This value is greater than 100%, which means there is likely an error in the actual yield provided. For the given amounts of reactants, the obtained yield of \(\mathrm{N}_{2}\) should not exceed the theoretical yield. Nonetheless, the calculated percent yield based on the given data is 107.7%.

Key Concepts

Balancing Chemical Equations

When dealing with chemical reactions, one essential step is balancing the equation. This ensures that the number of atoms for each element remains the same on both sides of the reaction. It's like making sure a recipe has the right amount of ingredients to match the finished dish. In our exercise, we started with \(\mathrm{NH}_{3} + \mathrm{NO} \rightarrow \mathrm{N}_{2} + \mathrm{H}_{2}\mathrm{O}\).

The equation wasn't balanced initially, so we adjusted coefficients. We changed the coefficient of ammonia \( (\mathrm{NH}_{3}) \) to 2, and water \((\mathrm{H}_{2}\mathrm{O})\) to 3. Now, the balanced equation is:* \(2\mathrm{NH}_{3}+\mathrm{NO}\rightarrow \mathrm{N}_{2}+3\mathrm{H}_{2}\mathrm{O}\)This balance ensures the law of conservation of mass is satisfied, meaning no atoms are lost or gained in the chemical process.

Stoichiometry

Stoichiometry is the calculation process of reactants and products in a chemical reaction. Think of it as the math behind chemistry, which helps determine how much of each substance is involved. In our balanced equation, stoichiometry tells us the ratio between the reactants and products.

For every 2 moles of \(\mathrm{NH}_{3}\), 1 mole of \(\mathrm{NO}\) is needed to produce 1 mole of \(\mathrm{N}_{2}\) and 3 moles of water. By knowing the amount of one reactant or product, we can use stoichiometry to calculate the amounts of others required or produced. It’s a vital tool for predicting quantities in chemical reactions.

Limiting Reactant

The limiting reactant is the substance that runs out first during a chemical reaction, limiting the amount of product formed. It’s like running out of flour when baking cookies; everything stops when a key ingredient is gone.

To find the limiting reactant, calculate the moles of each reactant. In our exercise, we had 0.881 moles of \(\mathrm{NH}_{3}\) and 0.733 moles of \(\mathrm{NO}\). By dividing these amounts by their coefficients from the balanced equation, you decide which one gives the smallest result.

• \(\frac{0.881}{2} = 0.441\)
• \(\frac{0.733}{1} = 0.733\)

With \(\mathrm{NH}_{3}\) having the smallest ratio, it is the limiting reactant, ending the reaction once it is consumed.

Percent Yield

Percent yield is a measure of the efficiency of a reaction, showing how much product was actually made versus what was theoretically possible. It’s calculated using:\[\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\]In the exercise, the theoretical yield of \(\mathrm{N}_{2}\) was calculated as 12.35 grams, but the actual yield was 13.3 grams. So, the percent yield was:

• \(\frac{13.3\, \text{g}}{12.35\, \text{g}} \times 100\% = 107.7\%\)

A yield over 100% usually indicates an error, as it suggests more product was obtained than possible. This kind of error may stem from measurement inaccuracies or impurities in the reactants.