Question

Silver sulfide, the tarnish on silverware, comes from reaction of silver metal with hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) : $$ \mathrm{Ag}+\mathrm{H}_{2} \mathrm{~S}+\mathrm{O}_{2} \longrightarrow \mathrm{Ag}_{2} \mathrm{~S}+\mathrm{H}_{2} \mathrm{O} \quad \text { Unbalanced } $$ (a) Balance the equation. (b) If the reaction were used intentionally to prepare \(\mathrm{Ag}_{2}\) S, how many grams would be formed from \(496 \mathrm{~g}\) of \(\mathrm{Ag}, 80.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\), and \(40.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\) if the reaction takes place in \(95 \%\) yield based on the limiting reactant?

Short answer

The balanced equation is \(2\mathrm{Ag} + \mathrm{H}_2\mathrm{S} + \mathrm{O}_2 \rightarrow \mathrm{Ag}_2\mathrm{S} + 2\mathrm{H}_2\mathrm{O}\). The mass of \(\mathrm{Ag}_2\mathrm{S}\) produced is 294.20 g.

Step-by-step solution

Write the unbalanced equation

The problem provides the unbalanced equation: \[\mathrm{Ag} + \mathrm{H}_2\mathrm{S} + \mathrm{O}_2 \rightarrow \mathrm{Ag}_2\mathrm{S} + \mathrm{H}_2\mathrm{O}\]

Balance the chemical equation

To balance the equation, let's first balance the silver (Ag) atoms by putting a coefficient of 2 in front of \(\mathrm{Ag}\):\[2\mathrm{Ag} + \mathrm{H}_2\mathrm{S} + \mathrm{O}_2 \rightarrow \mathrm{Ag}_2\mathrm{S} + \mathrm{H}_2\mathrm{O}\]Now, let's balance the oxygen atoms. Since there are 2 oxygen atoms on the reactant side, we need 2 hydrogen atoms on the product side. So, we change \(\mathrm{H}_2\mathrm{O}\) to 2:\(\mathrm{H}_2\mathrm{O}\):\[2\mathrm{Ag} + \mathrm{H}_2\mathrm{S} + \mathrm{O}_2 \rightarrow \mathrm{Ag}_2\mathrm{S} + 2\mathrm{H}_2\mathrm{O}\]This equation is now balanced with 2 Ag, 2 H, 1 S, and 2 O on both sides.

Calculate the molar masses

We need the molar masses of each reactant and product:- Molar mass of \(\mathrm{Ag}\) is \(107.87 \ \text{g/mol}\)- Molar mass of \(\mathrm{H}_2\mathrm{S}\) is \(34.08 \ \text{g/mol}\)- Molar mass of \(\mathrm{O}_2\) is \(32.00 \ \text{g/mol}\)- Molar mass of \(\mathrm{Ag}_2\mathrm{S}\) is \(247.74 \ \text{g/mol}\)

Determine the moles of each reactant

We compute the number of moles for each reactant:- Moles of \(\mathrm{Ag}\) = \(\frac{496 \ \text{g}}{107.87 \ \text{g/mol}} = 4.60\ \text{moles}\)- Moles of \(\mathrm{H}_2\mathrm{S}\) = \(\frac{80.0 \ \text{g}}{34.08 \ \text{g/mol}} = 2.35\ \text{moles}\)- Moles of \(\mathrm{O}_2\) = \(\frac{40.0 \ \text{g}}{32.00 \ \text{g/mol}} = 1.25\ \text{moles}\)

Identify the limiting reactant

Using the balanced equation (where 2 moles of Ag react with 1 mole of \(\mathrm{H}_2\mathrm{S}\) and 1 mole of \(\mathrm{O}_2\)), calculate the moles needed for each:- For 4.60 moles of \(\mathrm{Ag}\), you need 2.30 moles of \(\mathrm{H}_2\mathrm{S}\) and 2.30 moles of \(\mathrm{O}_2\)Since only 1.25 moles of \(\mathrm{O}_2\) are available, \(\mathrm{O}_2\) is the limiting reactant.

Calculate moles of \(\mathrm{Ag}_2\mathrm{S}\) produced

According to the balanced equation, 1 mole of \(\mathrm{O}_2\) produces 1 mole of \(\mathrm{Ag}_2\mathrm{S}\). Thus, 1.25 moles of \(\mathrm{O}_2\) will produce 1.25 moles of \(\mathrm{Ag}_2\mathrm{S}\).

Convert moles of \(\mathrm{Ag}_2\mathrm{S}\) to grams

Calculate the mass of \(\mathrm{Ag}_2\mathrm{S}\) produced using its molar mass (247.74 g/mol):\[1.25\ \text{moles} \times 247.74\ \text{g/mol} = 309.68\ \text{g}\]

Adjust for 95% yield

Since the reaction has a 95% yield, multiply the theoretical yield by 0.95:\[309.68\ \text{g} \times 0.95 = 294.20\ \text{g}\]

Key concepts

Stoichiometry

Stoichiometry is all about quantifying relationships in chemical reactions. By balancing the chemical equation, we ensure that atoms are neither created nor destroyed, fulfilling the law of conservation of mass. This means the number of atoms for each element on the reactant side equals that on the product side. When balancing, we adjust coefficients (the numbers before substances in the equation) to get it right. For example, in the equation given, we need two silver (Ag) atoms to react properly, leading to a balanced equation:

- 2Ag + H₂S + O₂ → Ag₂S + 2H₂O.

Why is this important? A balanced equation allows us to use stand-alone values like molar masses and mole ratios to calculate other quantities, such as the mass or volume of reactants and products involved in a reaction. These calculations rely on molar concepts: for instance, using the molar mass of a substance to convert grams to moles, facilitating a deeper understanding of the reaction's dynamics.

Limiting Reactant

Identifying the limiting reactant is essential in any stoichiometric problem. The limiting reactant entirely gets used up first in a reaction, determining the maximum amount of product formed.

Let's break down the process:

• Calculate the moles of each reactant using their given masses and molar masses.
• Use the balanced equation to understand how many moles of each reactant are required to react perfectly together.
• The reactant that produces the least amount of product is the limiting reactant.

In our case, even though we have enough silver and hydrogen sulfide, oxygen ( O₂) is our limiting reactant because we only have 1.25 moles, less than needed according to stoichiometric calculations. Thus, oxygen controls how much silver sulfide ( Ag₂S) can be produced, despite having enough of the other reactants. Recognizing the limiting reactant is crucial for determining actual product yield and in economizing the use of resources.

Percent Yield

Percent yield is a measure of reaction efficiency, indicating the percentage of the theoretical yield actually obtained. Real-world reactions rarely yield products perfectly due to practical losses or side reactions, which makes percent yield a practical metric.

Here's how it works:

• Theoretical yield is the maximum amount of product expected from the limiting reactant.
• Actual yield is the amount of product physically collected from the reaction.
• Percent yield is calculated as: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%% \]

In the given problem, after determining the theoretical mass of Ag₂S as 309.68 g, the calculation of the percent yield using a 95% factor adjusts this to 294.20 g. Being able to predict and understand percent yield helps in troubleshooting reactions and improving industrial efficiency.