Problem 98 - Chemistry and Chemical Reactivity

Question

A The elements silver, molybdenum, and sulfur combine to form AgaMoS., What is the maximum mass of \(\mathrm{Ag}_{2} \mathrm{MoS}_{4}\) that can be obtained if \(8.63 \mathrm{g}\) of silver, \(3.36 \mathrm{g}\) of molybdenum, and \(4.81 \mathrm{g}\) of sulfur are combined? (Hint: What is the limiting reactant?)

Step-by-Step Solution

Answer

The maximum mass of \(\text{Ag}_2\text{MoS}_4\) that can be obtained is 12.31 g.

Step 1: Write Chemical Formula and Molar Masses

The compound formed when silver, molybdenum, and sulfur combine is \(\text{Ag}_2\text{MoS}_4\). We need to find the molar masses of each element: Silver (Ag) = 107.87 g/mol, Molybdenum (Mo) = 95.95 g/mol, Sulfur (S) = 32.07 g/mol. Using these, calculate the molar mass of \(\text{Ag}_2\text{MoS}_4\):\[ 2 \times 107.87 + 95.95 + 4 \times 32.07 = 351.81 \text{ g/mol} \]

Step 2: Calculate Moles of Each Element

Calculate the number of moles of each element based on the given masses:- Moles of silver: \[ \frac{8.63 \text{ g}}{107.87 \text{ g/mol}} = 0.080 \text{ mol} \]- Moles of molybdenum: \[ \frac{3.36 \text{ g}}{95.95 \text{ g/mol}} = 0.035 \text{ mol} \]- Moles of sulfur: \[ \frac{4.81 \text{ g}}{32.07 \text{ g/mol}} = 0.150 \text{ mol} \]

Step 3: Identify Limiting Reactant

The chemical formula \(\text{Ag}_2\text{MoS}_4\) requires a 2:1:4 ratio of silver to molybdenum to sulfur. Check how many moles of \(\text{Ag}_2\text{MoS}_4\) each element can produce:- Silver can produce \( \frac{0.080}{2} = 0.040 \text{ mol} \) of \(\text{Ag}_2\text{MoS}_4\).- Molybdenum can produce \(0.035 \text{ mol}\).- Sulfur can produce \( \frac{0.150}{4} = 0.0375 \text{ mol} \).Molybdenum is the limiting reactant as it produces the least amount of \(\text{Ag}_2\text{MoS}_4\).

Step 4: Calculate Mass of Ag2MoS4 Produced

Using the moles of the limiting reactant (molybdenum), calculate the mass of \(\text{Ag}_2\text{MoS}_4\) that can be formed:\[ \text{Mass} = 0.035 \text{ mol} \times 351.81 \text{ g/mol} = 12.31 \text{ g} \]

Key Concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that determines how much product can ultimately be formed. This is because it is consumed first, thereby stopping the reaction. Identifying the limiting reactant involves calculating how much product each reactant could potentially produce and seeing which produces the least amount. This is key in stoichiometry, which is used to predict yields in chemical reactions.
For example, if you have three ingredients to make a cake, but one ingredient runs out first, you can't make any more cakes no matter how much of the other ingredients are left. Here, in the formation of \( \text{Ag}_2\text{MoS}_4 \), molybdenum is identified as the limiting reactant since it can form only \(0.035 \text{ mol}\) of product, less than what the other reactants could form.

Molar Mass Calculation

The molar mass of a compound is the sum of the masses of all its constituent atoms, and it's crucial for converting between mass and moles. To find the molar mass of \( \text{Ag}_2\text{MoS}_4 \), we add up the atomic masses:

• Silver (Ag), with a molar mass of \(107.87 \text{ g/mol}\), used twice
• Molybdenum (Mo), \(95.95 \text{ g/mol}\)
• Sulfur (S), \(32.07 \text{ g/mol}\), used four times

Calculating it: \[ 2 \times 107.87 + 95.95 + 4 \times 32.07 = 351.81 \text{ g/mol} \] Understanding molar masses helps to transition from quantities in grams to quantities in moles, an essential skill in stoichiometry.

Chemical Reaction Equations

Chemical reaction equations are like recipes for reactions, showing which substances react, how many molecules of each substance participate, and what products are formed. The balanced equation provides the ratio of moles needed for the reaction to occur completely.
In our exercise, silver (Ag), molybdenum (Mo), and sulfur (S) react to form \( \text{Ag}_2\text{MoS}_4 \). The ratio, as shown in the compound's formula, is 2 Ag: 1 Mo: 4 S. This specific ratio must be maintained just like in a recipe, ensuring all reactants are used efficiently and the reaction can proceed fully without any leftovers, except when limited by the limiting reactant.

Moles Calculation

Moles are a unit that counts particles, allowing chemists to predict how much product can form from a given amount of reactants. The transition from grams to moles uses molar masses, which acts as a bridge in calculations.
To calculate moles from a given mass:

• For silver: \( \frac{8.63 \text{ g}}{107.87 \text{ g/mol}} = 0.080 \text{ mol} \)
• For molybdenum: \( \frac{3.36 \text{ g}}{95.95 \text{ g/mol}} = 0.035 \text{ mol} \)
• For sulfur: \( \frac{4.81 \text{ g}}{32.07 \text{ g/mol}} = 0.150 \text{ mol} \)

These calculations are critical for determining both how much reactant is present initially and what amount of product can be produced. By understanding how to convert between mass and moles, you gain clarity over the entire chemical reaction process.