Problem 86 - Chemistry The Central Science In Si Units

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?

Step-by-Step Solution

Answer

2.64 g of sodium sulfide are produced.

Step 1: Write the Balanced Chemical Equation

Hydrogen sulfide (H_2S) reacts with sodium hydroxide (NaOH) to form sodium sulfide (Na_2S) and water (H_2O). The balanced equation is:\[ \text{H}_2\text{S} + 2\text{NaOH} \rightarrow \text{Na}_2\text{S} + 2\text{H}_2\text{O} \]

Step 2: Find the Molar Masses

Calculate the molar masses of the reactants and products:- Hydrogen sulfide (H_2S): \( 2(1.01) + 32.07 = 34.09 \ \text{g/mol} \)- Sodium hydroxide (NaOH): \( 22.99 + 16.00 + 1.01 = 40.00 \ \text{g/mol} \)- Sodium sulfide (Na_2S): \( 2(22.99) + 32.07 = 78.05 \ \text{g/mol} \)

Step 3: Determine the Limiting Reactant

Calculate the moles of each reactant:- Moles of H_2S: \( \frac{1.25}{34.09} = 0.0367 \ \text{mol} \)- Moles of NaOH: \( \frac{2.00}{40.00} = 0.0500 \ \text{mol} \)Since the reaction requires 2 moles of NaOH for each mole of H_2S, and we have more moles of NaOH compared to what’s needed for 0.0367 mol H_2S, H_2S is the limiting reactant.

Step 4: Calculate Theoretical Yield of Sodium Sulfide

The stoichiometry of the reaction shows that 1 mole of H_2S produces 1 mole of Na_2S. Therefore, 0.0367 mol of H_2S will produce 0.0367 mol of Na_2S.Calculate the mass of Na_2S:\[ \text{mass of } \text{Na}_2\text{S} = 0.0367 \times 78.05 = 2.864 \ \text{g} \]

Step 5: Adjust for Reaction Yield

Given that the reaction yield is 92.0%, adjust the theoretical yield to find the actual yield:\[ \text{Actual mass of } \text{Na}_2\text{S} = 2.864 \times 0.92 = 2.636 \ \text{g} \]

Key Concepts

Balanced Chemical Equation

A balanced chemical equation serves as a foundation for understanding chemical reactions. It shows how the reactants transform into products, ensuring that the number of atoms for each element is conserved.
For instance, in the reaction of hydrogen sulfide with sodium hydroxide:

• Hydrogen sulfide (\( ext{H}_2 ext{S}\)) and sodium hydroxide (\( ext{NaOH}\)) are reactants.
• Sodium sulfide (\( ext{Na}_2 ext{S}\)) and water (\( ext{H}_2 ext{O}\)) are products.
• The balanced equation is: \[\text{H}_2\text{S} + 2\text{NaOH} \rightarrow \text{Na}_2\text{S} + 2\text{H}_2\text{O}\]
• This equation shows that one mole of hydrogen sulfide reacts with two moles of sodium hydroxide to form one mole of sodium sulfide and two moles of water.

Having a balanced equation is crucial as it provides the exact ratios of reactants needed to form products, guiding us in predicting the amount of product produced.

Molar Mass Calculation

Calculating molar mass is a key step in determining the amount of a substance in grams. It is the mass of one mole of a substance, measured in grams per mole (\( ext{g/mol}\)), and is essential for converting between mass and moles.
Here’s how it works for the involved compounds:

• For hydrogen sulfide (\( ext{H}_2 ext{S}\)): \[2(1.01) + 32.07 = 34.09 \, \text{g/mol}\]
• For sodium hydroxide (\( ext{NaOH}\)): \[22.99 + 16.00 + 1.01 = 40.00 \, \text{g/mol}\]
• For sodium sulfide (\( ext{Na}_2 ext{S}\)): \[2(22.99) + 32.07 = 78.05 \, \text{g/mol}\]

These calculations provide the necessary data to translate mass into moles, which is particularly useful when following the stoichiometric ratios in a balanced chemical equation.

Theoretical Yield

The theoretical yield is the maximum possible amount of product that can be produced in a chemical reaction under ideal conditions. It's derived from the stoichiometry of the balanced equation.
Based on the equation \( ext{H}_2 ext{S} + 2 ext{NaOH} \rightarrow ext{Na}_2 ext{S} + 2 ext{H}_2 ext{O}\):

• One mole of \( ext{H}_2 ext{S}\) should yield one mole of \( ext{Na}_2 ext{S}\).
• If you start with 0.0367 moles of \( ext{H}_2 ext{S}\), you should expect to form the same 0.0367 moles of \( ext{Na}_2 ext{S}\).
• To calculate the theoretical mass, multiply the moles by the molar mass of sodium sulfide:\[\text{mass of } \text{Na}_2\text{S} = 0.0367 \times 78.05 = 2.864 \, \text{g}\]

Knowing the theoretical yield is essential for comparing with the actual yield obtained to determine reaction efficiency.

Reaction Yield

Reaction yield refers to the efficiency of a reaction in producing the desired product. It's important to compare how much product is actually made to what could be made theoretically.
For this reaction, given the 92.0% yield:

• Theoretical yield of \( ext{Na}_2 ext{S}\) is 2.864 grams.
• Actual yield can be calculated as 92% of the theoretical yield:\[\text{Actual mass of } \text{Na}_2\text{S} = 2.864 \times 0.92 = 2.636 \, \text{g}\]
• This signifies that only 92% of what theoretically could be produced, was actually formed.

Calculating the reaction yield helps in understanding not only the practical outcome of a chemical process but also in identifying potential losses or inefficiencies during the reaction.