Question

A \(20.27-\mathrm{g}\) sample of a metal carbonate \(\left(\mathrm{MCO}_{3}\right)\) is combined with \(500 \mathrm{~mL}\) of a \(1.00 \mathrm{M} \mathrm{HCl}\) solution. The excess \(\mathrm{HCl}\) acid is then neutralized by \(32.80 \mathrm{~mL}\) of \(0.588 \mathrm{M} \mathrm{NaOH}\). Identify \(\underline{\mathrm{M}}\)

Short answer

The metal M is Magnesium (Mg).

Step-by-step solution

Write the Reaction Equations

First, write down the balanced chemical equations for the reactions involved. The first reaction is the metal carbonate and hydrochloric acid (HCl) reacting: \[ \text{MCO}_3 + 2 \text{HCl} \rightarrow \text{MCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \] The second reaction is the neutralization of the remaining HCl by sodium hydroxide (NaOH): \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

Calculate Excess HCl

To find the amount of excess HCl, use the neutralization reaction. Calculate the moles of NaOH used:\[ \text{moles of NaOH} = 0.588 \frac{\text{mol}}{\text{L}} \times 0.0328 \text{ L} = 0.0192864 \text{ mol} \] Since the reaction ratio is 1:1, this is also the moles of HCl neutralized.

Calculate Initial Moles of HCl

Calculate the initial moles of HCl in the 500 mL solution:\[ \text{moles of HCl} = 1.00 \frac{\text{mol}}{\text{L}} \times 0.500 \text{ L} = 0.500 \text{ mol} \]

Calculate Moles of HCl Reacted with MCO3

Subtract the moles of excess HCl (neutralized by NaOH) from the initial moles of HCl:\[ 0.500 \text{ mol (initial HCl)} - 0.0192864 \text{ mol (excess HCl)} = 0.4807136 \text{ mol} \]

Calculate Moles of MCO3

From the stoichiometry of the reaction \( \text{MCO}_3 + 2 \text{HCl} \rightarrow \text{MCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \), two moles of HCl react with one mole of \( \text{MCO}_3 \).\[ \text{moles of MCO}_3 = \frac{0.4807136}{2} = 0.2403568 \text{ mol} \]

Determine Molar Mass of MCO3

Use the moles of \( \text{MCO}_3 \) to find the molar mass. The molar mass \(\mu\) is given by:\[ \mu = \frac{\text{mass of MCO}_3}{\text{moles of MCO}_3} = \frac{20.27 \text{ g}}{0.2403568 \text{ mol}} \approx 84.34 \text{ g/mol} \]

Identify Metal M

Given \( \text{MCO}_3 \) has a molar mass of approximately 84.34 g/mol, subtract the molar mass of \( \text{CO}_3 \) (i.e., 60.01 g/mol) to find the molar mass of \( \text{M} \): \[ \text{Molar mass of M} = 84.34 - 60.01 = 24.33 \text{ g/mol} \] This corresponds to the molar mass of magnesium (Mg). Therefore, \( \underline{\text{M}} \) is Magnesium (Mg).

Key concepts

Chemical Reaction Stoichiometry

In chemistry, reaction stoichiometry helps us calculate the amounts of reactants and products formed in a chemical reaction. It operates on the principle that a chemical reaction is a balanced mathematical representation of matter transformation. The coefficients in a balanced chemical equation indicate the ratio in which the chemical species react. For instance, in the reaction of a metal carbonate with hydrochloric acid (HCl): \[ \text{MCO}_3 + 2 \text{HCl} \rightarrow \text{MCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]we see that one mole of the metal carbonate reacts with two moles of HCl. This ratio helps determine how much product will form or how much reactant is needed by comparing the number of moles. Using stoichiometry is essential in solving problems that involve chemical reactions, such as the given exercise where we identify a metal in a metal carbonate compound.

Neutralization Reaction

Neutralization reactions occur when an acid and a base react to form water and a salt. In the exercise above, the excess hydrochloric acid was neutralized by sodium hydroxide (NaOH): \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] This equation indicates a one-to-one stoichiometric relationship, meaning one mole of HCl reacts with one mole of NaOH. By measuring how much NaOH is needed to neutralize the leftover HCl, you can determine how much HCl remained after the initial reaction with the metal carbonate. Such reactions are crucial for calculating the exact quantities of substances in chemistry, especially when dealing with excess reagents. This step is key in facilitating accurate quantitative analyses.

Moles Calculation

Calculating moles is a fundamental concept in chemistry that allows us to quantitatively analyze reactions. The mole is a standard unit of measurement used to express the amount of a chemical substance. It relates masses of different substances to their molecule counts. Using the chemical reaction equations, you can calculate the moles of each reactant and product based on their molar concentrations and volumes. To calculate the moles of hydrochloric acid initially present in the solution:\[ \text{moles of HCl} = 1.00 \text{ M} \times 0.500 \text{ L} = 0.500 \text{ moles} \] After determining how much HCl remained after reacting with the carbonate, the excess HCl was neutralized, giving further insight into the quantity of HCl initially reacted. These calculations play a vital role in allowing chemists to track the progress of reactions and determine unknown quantities such as the identity of the metal carbonate.

Molar Mass Determination

Molar mass is pivotal in identifying unknown compounds, especially in exercises involving empirical formulas. By calculating the moles of a substance involved in a chemical reaction, you can determine its molar mass if its mass is known. This involves the relation: \[ \text{Molar mass} (\mu) = \frac{\text{mass of sample}}{\text{moles of sample}} \] In the exercise, the mass of the metal carbonate was given as 20.27 g. By dividing this by the calculated moles of the metal carbonate, its molar mass was determined:\[ \mu = \frac{20.27 \text{ g}}{0.2403568 \text{ mol}} \approx 84.34 \text{ g/mol} \]Once the molar mass of the whole compound was found, subtracting the known molar mass of the \( \text{CO}_3 \) group allowed for the determination of the unknown metal, leading to its identification as magnesium. Calculating molar mass is indispensable in uncovering unknown substances through a systematic approach.