Question

A \(1.268 \mathrm{~g}\) sample of a metal carbonate \(\left(\mathrm{MCO}_{3}\right)\) was treated with \(100.00 \mathrm{~mL}\) of \(0.1083 \mathrm{M}\) sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\), yielding \(\mathrm{CO}_{2}\) gas and an aqueous solution of the metal sulfate \(\left(\mathrm{MSO}_{4}\right)\). The solution was boiled to remove all the dissolved \(\mathrm{CO}_{2}\) and was then titrated with \(0.1241\) M NaOH. A \(71.02 \mathrm{~mL}\) volume of \(\mathrm{NaOH}\) was required to neutralize the excess \(\mathrm{H}_{2} \mathrm{SO}_{4}\). (a) What is the identity of the metal M? (b) How many liters of \(\mathrm{CO}_{2}\) gas were produced if the density of \(\mathrm{CO}_{2}\) is \(1.799 \mathrm{~g} / \mathrm{L} ?\)

Short answer

(a) The metal M is Barium. (b) 0.157 L of CO2 gas was produced.

Step-by-step solution

Calculate Moles of H2SO4 Used

First, find the moles of sulfuric acid initially present: \[ \text{moles of } \mathrm{H}_2\mathrm{SO}_4 = 0.100 \text{ L} \times 0.1083 \text{ mol/L} \]This gives us:\[ 0.01083 \text{ mol of } \mathrm{H}_2\mathrm{SO}_4 \]

Calculate Moles of Excess H2SO4

Determine moles of NaOH used in the titration:\[ \text{moles of NaOH} = 0.07102 \text{ L} \times 0.1241 \text{ mol/L} \]\[ = 0.008817682 \text{ mol of NaOH} \]Since NaOH reacts with H2SO4 in a 2:1 ratio, moles of excess \( \mathrm{H}_2\mathrm{SO}_4 \) are:\[ 0.008817682 \div 2 = 0.004408841 \text{ mol of } \mathrm{H}_2\mathrm{SO}_4 \]

Calculate Moles of H2SO4 Reacted

Subtract the excess moles from the initial moles:\[ 0.01083 - 0.004408841 = 0.006421159 \text{ mol of } \mathrm{H}_2\mathrm{SO}_4 \]

Find Moles of CO2 Produced

Since sulfuric acid reacts with the metal carbonate in a 1:1 ratio to produce \( \mathrm{CO}_2 \), the moles of \( \mathrm{CO}_2 \) are equal to the moles of \( \mathrm{H}_2\mathrm{SO}_4 \) reacted:\[ 0.006421159 \text{ mol of } \mathrm{CO}_2 \]

Calculate Mass of CO2 Produced

Determine the mass of \( \mathrm{CO}_2 \):\[ \text{Mass} = \text{moles} \times \text{molar mass of } \mathrm{CO}_2 \]\[ = 0.006421159 \times 44.01 \text{ g/mol} \]\[ = 0.28256364 \text{ g of } \mathrm{CO}_2 \]

Calculate Volume of CO2 Produced

Use the density of \( \mathrm{CO}_2 \) to find the volume:\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]\[ = \frac{0.28256364 \text{ g}}{1.799 \text{ g/L}} \]\[ \approx 0.157 \text{ L of } \mathrm{CO}_2 \]

Find Atomic Weight of the Metal M

The balanced equation for metal carbonate indicates each mole of \( \mathrm{H}_2\mathrm{SO}_4 \) reacts with one mole of \( \mathrm{MCO}_3 \), so moles of metal carbonate:\[ 0.006421159 \text{ mol} \]Find the molar mass of \( \mathrm{MCO}_3 \):\[ \text{Molar mass} = \frac{1.268 \text{ g}}{0.006421159 \text{ mol}} \approx 197.45 \] Subtract molar mass of \( \mathrm{CO}_3 \):\[ \mathrm{M} = 197.45 - 60.01 \approx 137.44 \] This suggests that M is Barium \( \approx 137.33 \text{ g/mol} \).

Summarize the Identity of Metal M and CO2 Volume

Metal \( \mathrm{M} \) in the metal carbonate is Barium. The volume of \( \mathrm{CO}_2 \) gas produced is approximately \( 0.157 \text{ L} \).

Key concepts

Sulfuric Acid

Sulfuric acid, \( \mathrm{H}_{2}\mathrm{SO}_{4} \), is a highly versatile chemical compound. It plays a significant role in various industrial processes and everyday chemical reactions. When discussing its use in stoichiometry, especially in reactions involving metal carbonates, its potency as a strong acid is critical. It donates two protons per molecule, allowing it to react in a 1:1 molar ratio with many other compounds, such as metal carbonates.

In the context of this exercise, sulfuric acid reacts with barium carbonate to produce carbon dioxide, water, and a metal sulfate. Here, the balanced chemical equation is essential for proper stoichiometry calculations. The amount of sulfuric acid initially used can determine the expected amount of product, particularly when related to excess reactants. By understanding the initial concentration and volume of the acid, students can calculate the moles of reactant, which is crucial for proceeding with the stoichiometric calculations of metal carbonates.

Barium Carbonate

Barium carbonate, \( \mathrm{BaCO}_3 \), is a common inorganic compound that finds usage in various chemical applications, such as ceramics and glasses. When dealing with its reactivity, it's important to note that barium carbonate reacts with acids, like sulfuric acid, to form carbon dioxide gas, water, and a soluble barium sulfate.

The identification of this compound in the reaction with sulfuric acid involves stoichiometry, where understanding the 1:1 molar ratio in the reaction is crucial. This involves using the balanced equation to comprehend the transformation of reactants to products. In such reactions, sulfuric acid and barium carbonate will produce an equivalent volume of carbon dioxide gas, which helps determine the overall stoichiometry of the reaction and identify the metal involved.

Carbon Dioxide Gas Volume

In chemical reactions involving metal carbonates and acids, the production of carbon dioxide gas, \( \mathrm{CO}_2 \), is a clear indicator of a reaction taking place. Understanding how to calculate the volume of \( \mathrm{CO}_2 \) produced is crucial in stoichiometry.

The volume can be calculated if the mass or moles of \( \mathrm{CO}_2 \) are known by using the density of \( \mathrm{CO}_2 \), which typically is \( 1.799 \text{ g/L} \). In the given exercise, the moles of \( \mathrm{CO}_2 \) are derived from the moles of sulfuric acid that reacted, since they are directly proportional.

• First, calculate the moles of \( \mathrm{CO}_2 \) based on the balanced chemical equation and known data.
• Convert the moles to mass using the molar mass of \( \mathrm{CO}_2 \), which is \( 44.01 \text{ g/mol} \).
• Finally, divide the mass by the density to find the volume.

This method is a fundamental aspect of stoichiometry in gas production reactions, allowing students to visualize real-world applications of theoretical chemistry.

Molecular Weight Calculation

Molecular weight calculations are fundamental in identifying unknown substances in a chemical reaction. The objective is to determine the molar mass of a compound based on given data, which simplifies the identification process of an unknown metal involved in the reaction.

In the scenario provided, the molar mass of the metal carbonate \( MCO_3 \) can be determined by dividing the given mass of the compound by the moles of \( MCO_3 \) derived from the stoichiometric ratios.

• The molecular weight or molar mass of carbon dioxide \( \mathrm{CO}_3 \) is relatively known at \( 60.01 \text{ g/mol} \).
• Subtract this from the calculated molar mass of \( MCO_3 \) to find the molecular weight of the metal \( M \).
• An approximate match to known periodic table values can confirm the identity of the metal, such as barium with a molar mass of \( 137.33 \text{ g/mol} \).

This method aids students in piecing together data to define the metal's identity accurately, reinforcing their understanding of stoichiometry and chemical equations.