Question

You are given a solid mixture of \(\mathrm{NaNO}_{2}\) and \(\mathrm{NaCl}\) and are asked to analyze it for the amount of NaNO, present. To do so you allow the mixture to react with sulfamic acid, \(\mathrm{HSO}_{3} \mathrm{NH}_{2},\) in water according to the equation \(\mathrm{NaNO}_{2}(\mathrm{aq})+\mathrm{HSO}_{3} \mathrm{NH}_{2}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{NaHSO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{N}_{2}(\mathrm{g}) $$ What is the weight percentage of \(\mathrm{NaNO}_{2}\) in \(1.232 \mathrm{g}\) of the solid mixture if reaction with sulfamic acid produces \(295 \mathrm{mL}\) of \(\mathrm{N}_{2}\) gas with a pressure of \(713 \mathrm{mm} \mathrm{Hg}\) at \(21.0^{\circ} \mathrm{C} ?\)

Short answer

The weight percentage of \(\mathrm{NaNO}_{2}\) is approximately 21.6%.

Step-by-step solution

Understand the Chemical Reaction

The reaction given is \(\mathrm{NaNO}_{2}(\mathrm{aq}) + \mathrm{HSO}_{3} \mathrm{NH}_{2}(\mathrm{aq}) \rightarrow \mathrm{NaHSO}_{4}(\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O}(\ell) + \mathrm{N}_{2}(\mathrm{g})\). This is a stoichiometric reaction where \(1\) mole of \(\mathrm{NaNO}_{2}\) produces \(1\) mole of \(\mathrm{N}_{2}\). Therefore, determining the moles of \(\mathrm{N}_{2}\) will give us the moles of \(\mathrm{NaNO}_{2}\) initially present.

Use Ideal Gas Law

Calculate the moles of \(\mathrm{N}_{2}\) using the ideal gas law equation: \(PV = nRT\). First, convert the pressure from mm Hg to atm: \(P = \frac{713}{760} \text{ atm}\). Convert the temperature to Kelvin: \(T = 21.0 + 273.15 = 294.15 \text{ K}\). The volume is \(295\text{ mL} = 0.295\text{ L}\), and \(R = 0.0821\text{ L atm } \text{mol}^{-1}\text{K}^{-1}\). Using these, solve for \(n\) (moles of \(\mathrm{N}_{2}\)): \[n = \frac{PV}{RT} = \frac{(713/760) \times 0.295}{0.0821 \times 294.15}\].

Calculate Moles of NaNO₂

Since the stoichiometry of the reaction is 1:1, the moles of \(\mathrm{NaNO}_{2}\) are equal to the moles of \(\mathrm{N}_{2}\) calculated in the previous step.

Determine the Mass of NaNO₂

Use the number of moles of \(\mathrm{NaNO}_{2}\) and its molar mass (\(23 + 14 + 2(16) = 69\text{ g/mol}\)) to find the mass of \(\mathrm{NaNO}_{2}\). \(\text{Mass of } \mathrm{NaNO}_{2} = n \times 69\text{ g/mol}\).

Calculate the Weight Percentage

Using the mass of \(\mathrm{NaNO}_{2}\) found, calculate the weight percentage of \(\mathrm{NaNO}_{2}\) in the mixture: \[\text{Weight Percentage} = \left(\frac{\text{Mass of } \mathrm{NaNO}_{2}}{1.232} \right) \times 100\%\].

Key concepts

Ideal Gas Law

The Ideal Gas Law is an essential equation used in chemistry to relate the pressure, volume, temperature, and amount of gas. Its formula is \( PV = nRT \), where \( P \) stands for pressure, \( V \) for volume, \( n \) for the number of moles, \( R \) is the gas constant (0.0821 L atm/mol K), and \( T \) is temperature in Kelvin. In practice, you need to make sure units are consistent to apply this equation effectively:

• Pressure must be in atmospheres (atm).
• Volume in liters (L).
• Temperature in Kelvin (K).

For the exercise, the pressure was given in millimeters of mercury (mm Hg), so it was converted to atmospheres using the conversion \( 1 \, \text{atm} = 760 \, \text{mm Hg} \). The volume given in milliliters was converted to liters by dividing by 1000. These conversions allowed the calculation of moles of gas, which plays a critical role in determining other unknowns in stoichiometric problems.

Chemical Reactions

Chemical reactions, like the one in this exercise, are transformations where reactants change into products. This specific reaction involves \( \text{NaNO}_2 \) and \( \text{HSO}_3 \text{NH}_2 \), and it generates \( \text{NaHSO}_4 \), water, and nitrogen gas in a 1:1 molar ratio.Understanding stoichiometry, which deals with the quantitative aspects of chemical reactions, is crucial here. Given the balanced equation:\[ \mathrm{NaNO}_{2}(\mathrm{aq}) + \mathrm{HSO}_{3} \mathrm{NH}_{2}(\mathrm{aq}) \rightarrow \mathrm{NaHSO}_4(\mathrm{aq}) + \mathrm{H}_2 \mathrm{O}( ext{l}) + \mathrm{N}_{2}(\mathrm{g}) \]The stoichiometric coefficients tell us that 1 mole of \( \text{NaNO}_2 \) and 1 mole of \( \text{HSO}_3 \text{NH}_2 \) react completely to produce 1 mole of nitrogen gas (2).This implies that the calculation of \( \text{N}_2 \) produced will directly inform us about the amount of \( \text{NaNO}_2 \) reacted, allowing us to solve for the desired quantity in steps.

Mole Concept

The mole concept is a fundamental idea in chemistry that provides a bridge between the atomic scale and the macroscopic world. One mole of any substance contains Avogadro's number, \( 6.022 \times 10^{23} \), of units (atoms, molecules, etc.).In the given problem, the key task is to find how much \( \text{NaNO}_2 \) is present in the solid mixture. By using the Ideal Gas Law, we found the number of moles of \( \text{N}_2 \) gas produced, which due to the stoichiometry of the reaction, is equal to the moles of \( \text{NaNO}_2 \) initially present.Knowing the mole concept allows us to:

• Translate the number of moles into mass using the molar mass (for \( \text{NaNO}_2 \), it's 69 g/mol).
• Calculate weight percentage by comparing the mass of \( \text{NaNO}_2 \) to the total mass of the mixture.

This helps in quantitatively analyzing the composition of the sample, a vital skill in chemical problem-solving.