Question

Given that \(24.0 \mathrm{~mL}\) of \(0.170 \mathrm{M}\) sodium iodide reacts \(\mathrm{com}-\) pletely with \(0.209 \mathrm{M}\) mercury(II) nitrate solution accordir to the chemical reaction $$\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NaI}(a q) \longrightarrow \mathrm{HgI}_{2}(s)+\mathrm{NaNO}_{3}(a q)$$ Draw a stoichiometry concept map and calculate: (a) the volume of \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) solution that reacted (b) the mass of \(\mathrm{HgI}_{2}\) precipitate that was produced

Short answer

(a) 9.76 mL of Hg(NO3)2 reacted; (b) 0.927 g of HgI2 was produced.

Step-by-step solution

Write the balanced chemical equation

We need to confirm the balanced chemical equation for the reaction. The given reaction is \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}(aq) + 2\mathrm{NaI}(aq) \longrightarrow \mathrm{HgI}_{2}(s) + 2\mathrm{NaNO}_{3}(aq)\). This is already balanced with respect to both mass and charge.

Calculate moles of NaI

The concentration of sodium iodide (\(\mathrm{NaI}\)) solution is given as \(0.170\ \mathrm{M}\) with a volume of \(24.0\ \mathrm{mL}\). To find the moles of \(\mathrm{NaI}\), we convert volume to liters and use the formula: \(\text{moles} = \text{concentration} \times \text{volume (in liters)}\). Hence, \(\text{moles of NaI} = 0.170\ \mathrm{M} \times 0.0240\ \mathrm{L} = 0.00408\ \text{mol}\).

Use stoichiometry to find moles of Hg(NO3)2

From the balanced equation, 1 mole of \(\mathrm{Hg(NO_3)_2}\) reacts with 2 moles of \(\mathrm{NaI}\). We have \(0.00408\) moles of \(\mathrm{NaI}\), so we use the stoichiometric ratio \((2:1)\) to calculate the moles of \(\mathrm{Hg(NO_3)_2}\): \(\text{moles of Hg(NO}_3)_2 = \frac{0.00408}{2} = 0.00204\) moles.

Calculate volume of Hg(NO3)2 solution

The concentration of \(\mathrm{Hg(NO}_3)_2\) solution is \(0.209\ \mathrm{M}\). We use the formula \(\text{volume} = \frac{\text{moles}}{\text{concentration}}\). Therefore, the volume of \(\mathrm{Hg(NO}_3)_2\) = \(\frac{0.00204}{0.209} = 0.00976\ \mathrm{L} \text{ or } 9.76\ \mathrm{mL}\).

Calculate mass of HgI2

Each mole of \(\mathrm{HgI}_2\) corresponds to 1 mole of \(\mathrm{Hg(NO}_3)_2\) based on the equation. We have \(0.00204\) moles of \(\mathrm{HgI}_2\). Using the molar mass of \(\mathrm{HgI}_2\), which is approximately \(454.4\ \mathrm{g/mol}\), the mass is \(0.00204\ \text{mol} \times 454.4\ \mathrm{g/mol} = 0.927\ \mathrm{g}\).

Key concepts

Balanced Chemical Equation

A balanced chemical equation is essential to accurately perform calculations in stoichiometry. An equation is considered balanced when the number of atoms for each element is the same on both sides, meaning both mass and charge are conserved. For the given problem, the equation is: - \( \mathrm{Hg(NO_3)_2}(aq) + 2\mathrm{NaI}(aq) \rightarrow \mathrm{HgI_2}(s) + 2\mathrm{NaNO_3}(aq) \)- In this equation, one molecule of mercury(II) nitrate reacts with two molecules of sodium iodide, producing one molecule of mercury(II) iodide and two molecules of sodium nitrate.- To ensure the equation is balanced, count the number of each type of atom on each side of the reaction. Here, both sides have: - 1 mercury (Hg) - 2 iodine (I) - 2 sodium (Na) - 2 nitrate (NO3)- Understanding this balanced equation allows us to use correct stoichiometric ratios for calculations.

Molar Mass Calculation

The molar mass calculation is crucial for determining how much of a substance is involved in a reaction. Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).- To calculate the molar mass of a compound, sum up the molar masses of its constituent atoms.- Mercury (Hg) has a molar mass of approximately 200.59 g/mol.- Iodine (I) has a molar mass of about 126.90 g/mol.- For mercury(II) iodide (\(\mathrm{HgI_2}\)), the calculation is: \[ \text{Molar mass of HgI}_2 = 200.59 \, \mathrm{g/mol} + 2(126.90 \, \mathrm{g/mol}) = 454.4 \, \mathrm{g/mol} \]- Understanding the molar mass helps in subsequent calculations regarding the amount and mass of a precipitate formed during the reaction.

Precipitate Formation

Precipitate formation is a common outcome in chemical reactions, often representing the conversion from dissolved ions to an insoluble solid. This occurs when product compounds have low solubility in water.- During the reaction in question, mercury(II) iodide (\(\mathrm{HgI_2}\)) precipitates out as a solid, noted by the (s) in the equation.- As the reaction proceeds: - The aqueous ions of mercury (\(\mathrm{Hg^{2+}}\)) and iodide (\(\mathrm{I^-}\)) combine to form a crystalline solid.- Precipitation is visible as the formation of a different phase, indicating a chemical change.- Knowing precipitate formation helps you determine the amount of substance that remains in a solid state, critical for yield calculations.

Mole Concept

The mole concept is a foundational concept in chemistry that acts as a bridge between atomic scale dimensions and macroscopic measurements.- One mole is defined as \(6.022 \times 10^{23}\) entities, whether they are atoms, molecules, or ions.- It helps relate the amount of reactants to products during stoichiometric calculations.- For example, in this reaction, you start with the moles of \(\mathrm{NaI}\) calculated using the formula: \[ \text{moles} = \text{concentration (M)} \times \text{volume (L)} \]- Here, the moles of \(\mathrm{NaI}\) are 0.00408 mol, telling us how many formula units are involved in the reaction.- Using the stoichiometric ratio from the balanced equation (2:1), you can further determine the moles of other reactants and products, revealing the complete reaction scale.