A 1.268 g sample of a metal carbonate (MCO 3 ) was treated with 100.00 mL of 0.1083 M sulfuric a

step 1 First, the reaction is MCO3 equals plus H2 -S -O -4 gas, sorry, ques, forms MSO4 equest, plus CO

A 1.268 g sample of a metal carbonate (MCO 3 ) was treated...

A 1.268 g sample of a metal carbonate (MCO $_{3} )$ was treated with 100.00 $\mathrm{mL}$ of 0.1083 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 $\mathrm{M} \mathrm{NaOH}$ . A 71.02 $\mathrm{mL}$ volume of 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 $\mathrm{g} / \mathrm{L}$ ?

A 1.268 g sample of a metal carbonate (MCO $_{3} )$ was treated with 100.00 $\mathrm{mL}$ of 0.1083 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 $\mathrm{M} \mathrm{NaOH}$ . A 71.02 $\mathrm{mL}$ volume of 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 $\mathrm{g} / \mathrm{L}$ ?

Key Concepts

Molarity and Volume Relationships

Molarity is a measure of concentration expressed as moles of solute per liter of solution. In quantitative chemical analysis, it is essential to use the molarity and the volume of solutions to calculate the number of moles involved in reactions. This principle is applied both for the initial acid solution and the titrant, providing a basis for subsequent stoichiometric calculations.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions using the balanced chemical equation. It is used here to relate the moles of acid initially present, the moles of acid that reacted with the carbonate, and then to determine the molar amount of the metal carbonate sample. Understanding stoichiometric relationships is essential to deducing the identity of the unknown metal from its carbonate compound.

Gas Density and Conversion

Gas density, which relates the mass and volume of a gas under specific conditions, is used to convert the mass of a gas produced into its corresponding volume. In problems where a gas is a reaction product, knowing its density allows for calculation of the volume of gas produced, linking experimental data to theoretical predictions.

Acid?Base Titration

Acid–base titration is an analytical technique used to determine the concentration of an acid or base in a solution. In this method, a titrant of known concentration (in this case, NaOH) is added to react with the excess acid remaining after the carbonate reaction, allowing calculation of the moles of acid that were not consumed in the primary reaction. This information is crucial in back-calculating how much acid reacted with the sample.

Back Titration

Back titration is a strategy used when the analyte reacts slowly or when the end point is difficult to determine directly. Instead of measuring the reaction directly, an excess of a standard reagent (here, sulfuric acid) is added to react completely with the analyte (metal carbonate), and the leftover reagent is titrated. This method is especially useful when the reaction of the analyte with the reagent produces a volatile product or when the reaction conditions complicate direct titration.

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