/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 92 You are given a solid mixture of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You are given a solid mixture of \(\mathrm{NaNO}_{2}\) and \(\mathrm{NaCl}\) and are asked to analyze it for the amount of \(\mathrm{NaNO}_{2}\) present. To do so, you allow the mixture to react with sulfamic acid, HSO \(_{3} \mathrm{NH}_{2}\), in water according to the equation $$\begin{aligned} \mathrm{NaNO}_{2}(\mathrm{aq})+\mathrm{HSO}_{3} \mathrm{NH}_{2}(\mathrm{aq}) & \rightarrow \\ & \mathrm{NaHSO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{N}_{2}(\mathrm{g}) \end{aligned}$$ 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 dry \(\mathrm{N}_{2}\) gas with a pressure of \(713 \mathrm{mm}\) Hg at \(21.0^{10} \mathrm{C} ?\)

Short Answer

Expert verified
The weight percentage of \( \text{NaNO}_2 \) is approximately 67.76\%.

Step by step solution

01

Understand the Reaction

The reaction of \( \text{NaNO}_2 \) with sulfamic acid (\( \text{HSO}_3 \text{NH}_2 \)) produces \( \text{NaHSO}_4 \), \( \text{H}_2 \text{O} \), and \( \text{N}_2 \) gas. The balanced chemical equation helps us see that 1 mole of \( \text{NaNO}_2 \) produces 1 mole of \( \text{N}_2 \). This relationship is crucial for finding moles and mass of \( \text{NaNO}_2 \).
02

Use PV = nRT to Find Moles of N2

Calculate the moles of \( \text{N}_2 \) produced using the ideal gas law equation \( PV = nRT \). Convert the pressure from mm Hg to atm: \( 713 \text{ mm Hg} \times \frac{1 \text{ atm}}{760 \text{ mm Hg}} = 0.938 \text{ atm} \). The temperature in Kelvin is \( 21.0^\circ C + 273.15 = 294.15 \text{ K} \). The volume is \( 295 \text{ mL} = 0.295 \text{ L} \). The ideal gas constant \( R \) is \( 0.0821 \text{ L}\cdot\text{atm}/\text{K}\cdot\text{mol} \). Thus, \( n = \frac{PV}{RT} = \frac{0.938 \times 0.295}{0.0821 \times 294.15} \approx 0.0121 \text{ mol} \).
03

Calculate Mass of NaNO_2

Using the moles of \( \text{N}_2 \) (0.0121 mol), we know that the same number of moles of \( \text{NaNO}_2 \) was present. The molar mass of \( \text{NaNO}_2 \) is approximately 69.00 g/mol. Thus, the mass of \( \text{NaNO}_2 \) is \( 0.0121 \text{ mol} \times 69.00 \text{ g/mol} \approx 0.8349 \text{ g} \).
04

Calculate Weight Percentage

The weight percentage of \( \text{NaNO}_2 \) in the mixture is calculated by dividing the mass of \( \text{NaNO}_2 \) by the total mass of the solid mixture and multiplying by 100. So, \( \frac{0.8349 \text{ g}}{1.232 \text{ g}} \times 100 \approx 67.76\% \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry that describes the relationship between pressure, volume, temperature, and the amount of gas in moles. This law is expressed as \( PV = nRT \), where:
  • \( P \) stands for pressure
  • \( V \) is the volume
  • \( n \) represents the number of moles
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature in Kelvin
To solve problems using the ideal gas law, you need to ensure all units are converted appropriately. For pressure, you usually convert to atmospheres, volume to liters, and temperature to Kelvin. This helps maintain consistency with the constant \( R \), which is typically 0.0821 L·atm/K·mol. By using the given conditions such as volume, pressure, and temperature, you can determine the number of moles of gas, which is crucial for understanding reactions involving gases.
Balanced Chemical Equation
A balanced chemical equation is essential for understanding the relationships between reactants and products in a reaction. It shows the molecules involved in a reaction with the correct stoichiometric coefficients. These coefficients ensure that the law of conservation of mass is upheld. This means the number of each type of atom is the same on both sides of the equation.
In the reaction \( \mathrm{NaNO}_{2}( ext{aq}) + \mathrm{HSO}_{3} \mathrm{NH}_{2}( ext{aq}) \rightarrow \mathrm{NaHSO}_{4}( ext{aq}) + \mathrm{H}_{2}\mathrm{O}( ext{l}) + \mathrm{N}_{2}( ext{g})\), the equation tells us:
  • 1 mole of \( \mathrm{NaNO}_{2} \) reacts with 1 mole of \( \mathrm{HSO}_{3} \mathrm{NH}_{2} \)
  • This produces 1 mole of \( \mathrm{N}_{2} \) gas
This balance is crucial for calculating how much product will form from a given amount of reactant. In this context, it helps determine the mass of \( \mathrm{NaNO}_{2} \) needed based on the moles of \( \mathrm{N}_{2} \) gas produced.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products. They can be complex, but they all follow the basic principle of rearranging atoms to form new substances. The specific reaction in this exercise involves sodium nitrite (\( \text{NaNO}_{2} \)) and sulfamic acid (\( \text{HSO}_{3} \text{NH}_{2} \)).
  • This reaction produces nitrogen gas (\( \text{N}_2 \)), water (\( \text{H}_2\text{O} \)), and sodium bisulfate (\( \text{NaHSO}_{4} \))
  • The generation of \( \text{N}_2 \) gas is significant because it allows the use of the ideal gas law to find the amount of \( \text{NaNO}_{2} \)
Understanding these transformations is fundamental. It allows chemists to predict the outcomes of a reaction and calculate how much of each substance is involved or produced. Reactions are often used in industry and laboratory settings to produce desired products efficiently.
Mole Calculations
Calculating moles is a central skill in chemistry. Moles allow chemists to quantify the amount of a substance, akin to counting atoms or molecules considering Avogadro's number (approximately \(6.022 \times 10^{23}\)). In the context of the given problem, we can determine the moles of \( \text{N}_2 \) gas produced by using the ideal gas law.
Once the moles of \( \text{N}_2 \) are known, you can infer the moles of \( \text{NaNO}_2 \) used. Given the balanced equation states a 1:1 ratio of \( \text{NaNO}_2 \) to \( \text{N}_2 \), the same number of moles applies to both compounds, facilitating further calculations.
  • To convert moles to grams, simply multiply the moles by the molar mass of the compound (e.g., \( \text{NaNO}_2 = 69.00 \text{ g/mol} \))
  • This conversion is vital for finding the mass composition of the mixture
These calculations help identify the mass percentage of substances in mixtures, providing crucial information for analyses and formulations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A helium-filled balloon of the type used in longdistance flying contains \(420,000 \mathrm{ft}^{3}\left(1.2 \times 10^{7} \mathrm{L}\right)\) of helium. Suppose you fill the balloon with helium on the ground, where the pressure is 737 mm \(\mathrm{Hg}\) and the temperature is \(16.0^{\circ} \mathrm{C}\). When the balloon ascends to a height of 2 miles, where the pressure is only \(600 .\) mm Hg and the temperature is \(-33^{\circ} \mathrm{C},\) what volume is occupied by the helium gas? Assume the pressure inside the balloon matches the external pressure. Comment on the result.

A xenon fluoride can be prepared by heating a mixture of \(\mathrm{Xe}\) and \(\mathrm{F}_{2}\) gases to a high temperature in a pressureproof container. Assume that xenon gas was added to a \(0.25-\mathrm{L}\) container until its pressure reached \(0.12 \mathrm{atm}\) at \(0.0^{\circ} \mathrm{C} .\) Fluorine gas was then added until the total pressure reached 0.72 atm at \(0.0^{\circ} \mathrm{C}\). After the reaction was complete, the xenon was consumed completely, and the pressure of the \(\mathrm{F}_{2}\) remaining in the container was 0.36 atm at \(0.0^{\circ} \mathrm{C} .\) What is the empirical formula of the xenon fluoride?

Nitrogen trifluoride is prepared by the reaction of ammonia and fluorine. $$ 4 \mathrm{NH}_{3}(\mathrm{g})+3 \mathrm{F}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{NH}_{4} \mathrm{F}(\mathrm{s})+\mathrm{NF}_{3}(\mathrm{g}) $$ If you mix \(\mathrm{NH}_{3}\) with \(\mathrm{F}_{2}\) in the correct stoichiometric ratio, and if the total pressure of the mixture is \(120 \mathrm{mm} \mathrm{Hg}\), what are the partial pressures of \(\mathrm{NH}_{3}\) and \(\mathrm{F}_{2} ?\) When the reactants have been completely consumed, what is the total pressure in the flask? (Assume \(T\) is constant.)

A compound containing \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O}\) is burned in excess oxygen. The gases produced by burning \(0.1152 \mathrm{g}\) are first treated to convert the nitrogen-containing product gases into \(\mathrm{N}_{2},\) and then the resulting mixture of \(\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O}, \mathrm{N}_{2},\) and excess \(\mathrm{O}_{2}\) is passed through a bed of \(\mathrm{CaCl}_{2}\) to absorb the water. The \(\mathrm{CaCl}_{2}\) increases in mass by \(0.09912 \mathrm{g} .\) The remaining gases are bubbled into water to form \(\mathrm{H}_{2} \mathrm{CO}_{3},\) and this solution is titrated with \(0.3283 \mathrm{M} \mathrm{NaOH} ; 28.81 \mathrm{mL}\) is required to achieve the second equivalence point. The excess \(\mathbf{O}_{2}\) gas is removed by reaction with copper metal (to give CuO). Finally, the \(\mathrm{N}_{2}\) gas is collected in a 225.0 -mL. flask, where it has a pressure of \(65.12 \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C} .\) In a separate experiment, the unknown compound is found to have a molar mass of \(150 \mathrm{g} / \mathrm{mol}\). What are the empirical and molecular formulas of the unknown compound?

A You have a \(550 .\) -mL. tank of gas with a pressure of 1.56 atm at \(24^{\circ} \mathrm{C} .\) You thought the gas was pure carbon monoxide gas, CO, but you later found it was contaminated by small quantities of gaseous \(\mathrm{CO}_{2}\) and \(\mathrm{O}_{2}\). Analysis shows that the tank pressure is 1.34 atm (at \(24^{\circ} \mathrm{C}\) ) if the \(\mathrm{CO}_{2}\) is removed. Another experiment shows that \(0.0870 \mathrm{g}\) of \(\mathrm{O}_{2}\) can be removed chemically. What are the masses of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) in the tank, and what is the partial pressure of each of the three gases at \(25^{\circ} \mathrm{C} ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.