Question

When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water. How many grams of sodium sulfide are formed if \(1.25 \mathrm{~g}\) of hydrogen sulfide is bubbled into a solution containing \(2.00 \mathrm{~g}\) of sodium hydroxide, assuming that the sodium sulfide is made in \(92.0 \%\) yield?

Short answer

In this reaction, hydrogen sulfide is the limiting reactant, with a 92.0% yield. We can calculate the actual mass of sodium sulfide formed, which is approximately 2.63 g.

Step-by-step solution

1. Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between hydrogen sulfide gas and sodium hydroxide is: \[H_2S(g) + 2 NaOH(aq) \rightarrow Na_2S(aq) + 2 H_2O(l)\]

2. Identify the information provided and the variable needed

We are given the mass of hydrogen sulfide (\(1.25 \ \text{g}\)) and the mass of sodium hydroxide (\(2.00 \ \text{g}\)). The percentage yield (\(92.0\%\)) helps to find the actual mass of sodium sulfide produced. We are looking for the mass of sodium sulfide formed during the reaction.

3. Calculate moles of reactants using their molecular weight

First, we need to determine the moles of each reactant. To do that, we will use the molar masses of hydrogen sulfide and sodium hydroxide. Molar mass of hydrogen sulfide (\(H_2S\)) = 1(2.016 g/mol) + 32.07 g/mol = 34.08 g/mol Molar mass of sodium hydroxide (\(NaOH\)) = 22.99 g/mol + 15.999 g/mol + 1.008 g/mol = 39.998 g/mol Now, calculate the moles of each reactant: Moles of hydrogen sulfide = mass/molar mass = \(\frac{1.25 \ \text{g}}{34.08 \ \text{g/mol}} = 0.0366 \ \text{mol}\) Moles of sodium hydroxide= mass/molar mass = \(\frac{2.00 \ \text{g}}{39.998 \ \text{g/mol}} = 0.0500 \ \text{mol}\)

4. Perform stoichiometric calculations to find limiting reactant

Next, we need to determine which reactant is limiting. Divide the moles of each reactant by their respective stoichiometric coefficients: \(0.0183 \ \text{mol} = \frac[]{0.0366 \ \text{mol}}{2}\), moles of hydrogen sulfide per 1 mole of sodium sulfide \(0.02500 \ \text{mol} = \frac[]{0.0500 \ \text{mol}}{2}\), moles of sodium hydroxide per 1 mole of sodium sulfide Since 0.0183 < 0.0250, hydrogen sulfide is the limiting reactant.

5. Calculate the theoretical mass of sodium sulfide

Using the moles of hydrogen sulfide as the limiting reactant, calculate the theoretical mass of sodium sulfide produced: Molar mass of sodium sulfide (\(Na_2S\)) = 2(22.99 g/mol) + 32.07 g/mol = 78.06 g/mol Moles of sodium sulfide = moles of hydrogen sulfide = 0.0366 mol Mass of sodium sulfide (theoretical) = moles \(\times\) molar mass = \(0.0366\ \text{mol}\) \(\times\) \(\frac[]{78.06 \ \text{g}}{1 \ \text{mol}}\) = 2.86 g

6. Calculate the actual mass of sodium sulfide using percentage yield

Finally, using the percentage yield value, we can calculate the actual mass of sodium sulfide formed: Mass of sodium sulfide (actual) = theoretical mass \(\times\) yield = \(2.86 \ \text{g}\) × \(0.9200\) = 2.63 g Hence, the mass of sodium sulfide produced during the reaction is approximately 2.63 g.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that determines the amount of product that can be formed. It limits the reaction because it gets consumed first, and once it's gone, the reaction stops.
In the problem we're discussing, hydrogen sulfide (\(H_2S\)) acts as the limiting reactant. When comparing molar amounts to the coefficients from the balanced equation, we find that hydrogen sulfide provides fewer moles of product compared to sodium hydroxide. Thus, it restricts the formation of sodium sulfide.
To pinpoint the limiting reactant:

• Convert the mass of each reactant to moles using their molar mass.
• Divide the number of moles by the coefficient from the balanced equation.
• The reactant that produces the least amount of product is the limiting reactant.

Understanding the limiting reactant concept is crucial for accurately predicting how much product your reaction will yield.

Percentage Yield

Once a reaction is complete, the percentage yield tells us how close we got to the maximum possible ("theoretical") yield. It's a measure of the reaction's efficiency.
The percentage yield is calculated using the formula:\[\text{Percentage Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]In this exercise, the percentage yield of sodium sulfide is given as 92.0%.
What does that mean? It indicates that only 92% of the theoretical amount of sodium sulfide was actually produced.
To find the actual mass of sodium sulfide, we multiply the theoretical yield by the percentage yield:

• If the theoretical yield is 2.86 g, then the actual yield is calculated as 2.86 g \(\times\) 0.9200, resulting in approximately 2.63 g.

Understanding percentage yield is vital for practical applications where resource efficiency matters.

Balanced Chemical Equation

A balanced chemical equation accurately represents the transformation of reactants into products. It respects the law of conservation of mass, stating that matter is neither created nor destroyed.
In our example, the balanced equation is:\[H_2S(g) + 2 NaOH(aq) \rightarrow Na_2S(aq) + 2 H_2O(l)\]This equation tells us:

• 1 molecule of hydrogen sulfide reacts with 2 molecules of sodium hydroxide to produce 1 molecule of sodium sulfide and 2 molecules of water.
• The coefficients (1, 2, 1, and 2, respectively) ensure that each element has the same number of atoms on both sides of the equation, keeping the equation balanced.

Knowing how to balance chemical equations is essential for performing stoichiometric calculations and understanding the proportions in which chemicals interact.

Molar Mass

Molar mass is the mass of one mole of a given substance. It's a critical concept for converting between grams and moles, a primary step in stoichiometry.
For our compounds, the molar masses are:

• Hydrogen sulfide (\(H_2S\)): 34.08 g/mol
• Sodium hydroxide (\(NaOH\)): 39.998 g/mol
• Sodium sulfide (\(Na_2S\)): 78.06 g/mol

To find the moles from a given mass:

• Use the formula \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).

Molar mass enables you to transition from the macroscopic world of grams to the microscopic world of molecules, atoms, and moles, helping you perform the necessary calculations to solve stoichiometry problems.