Problem 96 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

An amount of \(1.0 \times 10^{-3}\) moles of \(\mathrm{Ag}^{+}\) and \(1.0 \times 10^{-3}\) moles of \(\mathrm{CrO}_{4}^{2-}\) reacts together to form solid \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). What is the amount of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) formed? \((\mathrm{Ag}=108, \mathrm{Cr}=52)\) (a) \(0.332 \mathrm{~g}\) (b) \(0.166 \mathrm{~g}\) (c) \(332 \mathrm{~g}\) (d) \(166 \mathrm{~g}\)

Step-by-Step Solution

Answer

0.166 g

Step 1 - Write the balanced chemical equation

Determine the balanced chemical equation for the reaction of silver ions (\text{Ag}^+) and chromate ions (\text{CrO}_4^{2-}) to form solid silver chromate (\text{Ag}_2\text{CrO}_4). The balanced equation is: \[ 2\text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq) \rightarrow \text{Ag}_2\text{CrO}_4 (s). \]

Step 2 - Determine the limiting reactant

With equal moles of \text{Ag}^+ and \text{CrO}_4^{2-} and the stoichiometry of the reaction being 2:1, \text{Ag}^+ is the limiting reactant as 2 moles of \text{Ag}^+ are required for every 1 mole of \text{CrO}_4^{2-}. Thus, the number of moles of \text{Ag}_2\text{CrO}_4 formed will be half the moles of \text{Ag}^+ ions, which is: \[ 0.5 \times 1.0 \times 10^{-3} = 0.5 \times 10^{-3} \text{moles of} \text{Ag}_2\text{CrO}_4. \]

Step 3 - Calculate the molar mass of \text{Ag}_2\text{CrO}_4

Find the molar mass of \text{Ag}_2\text{CrO}_4 by summing the atomic masses of 2 silver (Ag) and 1 chromium (Cr) and 4 oxygen (O) atoms: \[ \text{Molar mass of} \text{Ag}_2 \text{CrO}_4 = 2 \times 108 + 52 + 4 \times 16 = 332 \text{g/mol}. \]

Step 4 - Calculate the mass of \text{Ag}_2\text{CrO}_4 formed

Multiply the moles of \text{Ag}_2\text{CrO}_4 by its molar mass to get the mass of the precipitate formed: \[ \text{Mass of} \text{Ag}_2\text{CrO}_4 = 0.5 \times 10^{-3} \text{moles} \times 332 \text{g/mol} = 0.166 \text{g}. \]

Key Concepts

Limiting Reactant Calculation

Understanding the concept of the limiting reactant is essential when solving stoichiometry problems. In a chemical reaction, the limiting reactant, also known as the limiting reagent, is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reactant, since the reaction cannot proceed without it.

When analyzing a reaction to identify the limiting reactant, you start by comparing the mole ratio of the reactants with the ratio in the balanced chemical equation. If the moles of a reactant divided by its coefficient in the balanced equation is the smallest compared to all other reactants, then that reactant is the limiting one.

In the given exercise, with the balanced chemical equation being \( 2\text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq) \rightarrow \text{Ag}_2\text{CrO}_4 (s) \), and given equal moles of \( \text{Ag}^+ \) and \( \text{CrO}_4^{2-} \), it is evident that \( \text{Ag}^+ \) is the limiting reactant because you need two \( \text{Ag}^+ \) ions for every one \( \text{CrO}_4^{2-} \) ion. Once identified, we can then use the amount of the limiting reactant to calculate the amount of product formed.

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in chemistry. Balanced equations reflect the conservation of mass and the principle that atoms cannot be created or destroyed in a chemical reaction.

To balance an equation, firstly, write the unbalanced equation with the reactants on the left and the products on the right. Then systematically adjust the coefficients, which are the numbers in front of compounds or elements, to ensure the same number of atoms of each element appears on both sides of the equation.

Take the example of \( \text{Ag}^+ \) and \( \text{CrO}_4^{2-} \) reacting to form \( \text{Ag}_2\text{CrO}_4 \). We start with the skeletal equation:\text{Ag}^+ + \text{CrO}_4^{2-} \rightarrow \text{Ag}_2\text{CrO}_4\
Reviewing each element, we notice that silver (\(\text{Ag}\)) is not balanced because there are two silver atoms in the product but only one in the reactants. As such, we adjust the coefficient of \(\text{Ag}^+\) to 2, which balances the silver atoms and gives us the balanced chemical equation:\2\text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq) \rightarrow \text{Ag}_2\text{CrO}_4 (s)\. This equation now matches the principle of conservation of mass.

Molar Mass Determination

Molar mass is a critical value in stoichiometry which represents the mass of one mole of a substance. To calculate the molar mass, we sum up the atomic masses of all the atoms present in the formula of the substance. The atomic masses can be found on the periodic table and are usually given in atomic mass units (amu) or grams per mole (g/mol).

For the silver chromate (\( \text{Ag}_2\text{CrO}_4 \) ) formed in our exercise, we determine its molar mass by adding together the atomic masses of two silver atoms, one chromium atom, and four oxygen atoms:

\[ \text{Molar mass of } \text{Ag}_2 \text{CrO}_4 = 2 \times \text{Atomic mass of Ag} + \text{Atomic mass of Cr} + 4 \times \text{Atomic mass of O} \]
Using the given atomic masses, we get:

\[ \text{Molar mass of } \text{Ag}_2 \text{CrO}_4 = 2 \times 108 + 52 + 4 \times 16 = 332 \text{g/mol} \]
This step is fundamental because the molar mass allows us to convert between the mass of a substance and the number of moles, a vital step in stoichiometry for predicting how much product can be formed from a given amount of reactants.