Question

A quantity of \(10 \mathrm{~g}\) of a piece of marble was put into excess of dilute \(\mathrm{HCl}\) acid. When the reaction was complete, \(1120 \mathrm{~cm}^{3}\) of \(\mathrm{CO}_{2}\) was obtained at \(0^{\circ} \mathrm{C}\) and 1 atm. The percentage of \(\mathrm{CaCO}_{3}\) in the marble is (a) \(5 \%\) (b) \(25 \%\) (c) \(50 \%\) (d) \(2.5 \%\)

Short answer

The percentage of \text{CaCO}_3 in the marble is 50%.

Step-by-step solution

Write down the chemical reaction

The reaction between calcium carbonate (\text{CaCO}_3) and hydrochloric acid (\text{HCl}) produces calcium chloride (\text{CaCl}_2), carbon dioxide (\text{CO}_2), and water (\text{H}_2\text{O}). The balanced chemical equation is: \[ \text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l) \]

Calculate the molar volume of a gas at STP

At Standard Temperature and Pressure (STP), which is 0°C (273 K) and 1 atm, one mole of an ideal gas occupies a volume of 22.4 liters (or 22400 cm³).

Calculate the moles of \text{CO}_2 produced

Use the volume of \text{CO}_2 produced and the molar volume at STP to calculate the moles of \text{CO}_2. The formula to use is: \[ n = \frac{V}{V_m} \] where \(n\) is the number of moles, \(V\) is the volume of gas at STP, and \(V_m\) is the molar volume at STP. So, \[ n = \frac{1120 \text{ cm}^3}{22400 \text{ cm}^3/mol} = 0.05 \text{ mol} \]

Calculate the mass of \text{CaCO}_3 that reacted

From the balanced chemical equation, 1 mole of \text{CaCO}_3 produces 1 mole of \text{CO}_2. So, if 0.05 moles of \text{CO}_2 is produced, it means that 0.05 moles of \text{CaCO}_3 reacted. Calculate the mass using the molar mass of \text{CaCO}_3 (\text{Ca} = 40, \text{C} = 12, \text{O} = 16 x 3 = 48, total = 100 g/mol): \[ m = n \times M \] where \(m\) is mass, \(n\) is moles, and \(M\) is molar mass. \[ m = 0.05 \text{ mol} \times 100 \text{ g/mol} = 5 \text{ g} \]

Calculate the percentage purity

Calculate the percentage of \text{CaCO}_3 in the marble by comparing the mass of \text{CaCO}_3 that reacted with the initial mass of the marble sample. Use the formula: \[ \text{Percentage} = \left( \frac{\text{Mass of } \text{CaCO}_3}{\text{Initial mass of marble}} \right) \times 100\text{%} \] \[ \text{Percentage} = \left( \frac{5 \text{ g}}{10 \text{ g}} \right) \times 100\text{%} = 50\text{%} \]

Key concepts

Understanding Chemical Reaction Equations

At the heart of stoichiometry lies the ability to interpret chemical reaction equations. A balanced equation provides a snapshot of the conversion of reactants into products, showing the exact number of particles involved on a molecular level. Consider the reaction of calcium carbonate with hydrochloric acid; the equation is balanced when it accurately reflects the preservation of mass and the law of conservation of energy.

Each reactant and product is written with a specific stoichiometry, indicating the molar ratios in which they react and are formed. Understanding these ratios is crucial as they are the foundation for all subsequent stoichiometric calculations. In our example, the 1:1 ratio between calcium carbonate and carbon dioxide informs us that each mole of calcium carbonate gives rise to one mole of carbon dioxide. Grasping this concept allows students to move beyond the mere memorization of equations and to start applying them in practical contexts like calculating reactant or product quantities.

Molar Volume at STP: A Cornerstone Concept

The molar volume of a gas at Standard Temperature and Pressure (STP), which is 0°C and 1 atm, is essential for solving stoichiometry problems involving gases. At these conditions, one mole of any ideal gas occupies 22.4 liters, or 22400 cm³. This concept is pivotal because it provides a direct link between the volume of gas observed experimentally and the amount of substance in moles.

Using the molar volume allows students to perform conversions between the physical volume of a gas and the number of moles, enabling them to calculate the stoichiometry of reactions involving gaseous products or reactants. For example, given a volume of carbon dioxide gas produced in a reaction, one can calculate the moles of carbon dioxide and, by extension, the moles of the reactant consumed, assuming ideal gas behavior.

Calculating Percentage Purity

The final step in handling stoichiometry problems often involves deducing the purity of a sample. Percentage purity calculations assess the quality of a reactant by comparing the mass of the pure substance to the mass of the impure sample. In our original exercise, the impure sample was a piece of marble primarily composed of calcium carbonate.

The formula to calculate purity is straightforward: \[ \text{Percentage purity} = \frac{\text{Mass of pure substance}}{\text{Total mass of sample}} \times 100\text{%} \]
The percentage indicates what portion of the total mass is the desired substance. This is particularly important in reactions where impurities could affect the outcome, such as in pharmaceutical synthesis or material manufacturing. Through this calculation, students can connect stoichiometric principles with tangible measures of chemical quality.