/*! 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 26 At higher temperatures, NaHCO is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At higher temperatures, NaHCO is converted quantitatively to \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) $$ 2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ Heating a \(1.7184-\mathrm{g}\) sample of impure NaHCO \(_{3}\) gives \(0.196 \mathrm{g}\) of \(\mathrm{CO}_{2} .\) What was the mass percent of \(\mathrm{NaHCO}_{3}\) in the original 1.7184 -g sample?

Short Answer

Expert verified
The mass percent of \( \mathrm{NaHCO}_{3} \) is approximately 43.47\%.

Step by step solution

01

Calculate Moles of \( \mathrm{CO}_{2} \) Produced

The molar mass of \( \mathrm{CO}_{2} \) is approximately 44.01 g/mol. Calculate the moles of \( \mathrm{CO}_{2} \) using the mass given:\[\text{Moles of } \mathrm{CO}_{2} = \frac{0.196 \text{ g}}{44.01 \text{ g/mol}} \approx 0.00445 \text{ mol}\]
02

Relate \( \mathrm{CO}_{2} \) to \( \mathrm{NaHCO}_{3} \)

The balanced chemical equation shows that 2 moles of \( \mathrm{NaHCO}_{3} \) produce 1 mole of \( \mathrm{CO}_{2} \). Therefore, the moles of \( \mathrm{NaHCO}_{3} \) that reacted are:\[\text{Moles of } \mathrm{NaHCO}_{3} = 2 \times 0.00445 \text{ mol} = 0.0089 \text{ mol}\]
03

Find Mass of \( \mathrm{NaHCO}_{3} \) That Reacted

The molar mass of \( \mathrm{NaHCO}_{3} \) is approximately 84.01 g/mol. Calculate the mass of \( \mathrm{NaHCO}_{3} \) that reacted:\[\text{Mass of } \mathrm{NaHCO}_{3} = 0.0089 \text{ mol} \times 84.01 \text{ g/mol} \approx 0.747 \text{ g}\]
04

Determine Mass Percent of \( \mathrm{NaHCO}_{3} \)

Calculate the mass percent of \( \mathrm{NaHCO}_{3} \) in the original sample using the mass of \( \mathrm{NaHCO}_{3} \) that reacted:\[\text{Mass percent of } \mathrm{NaHCO}_{3} = \left(\frac{0.747 \text{ g}}{1.7184 \text{ g}}\right) \times 100\% \approx 43.47\%\]

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.

chemical reactions
In chemistry, a chemical reaction is a process where substances, known as reactants, transform into different substances, called products. This transformation involves breaking of bonds in reactants and formation of new bonds in products. For instance, when heating sodium bicarbonate (\(\mathrm{NaHCO}_3\)), it decomposes into sodium carbonate (\(\mathrm{Na}_2\mathrm{CO}_3\)), carbon dioxide (\(\mathrm{CO}_2\)), and water (\(\mathrm{H}_2\mathrm{O}\)). During this process, energy is absorbed to break the bonds, and released when new ones are formed.

Chemical reactions can be influenced by various factors such as temperature, concentration of reactants, and even the presence of catalysts. Increasing the temperature generally increases the rate of a chemical reaction, allowing atoms to collide more energetically, leading to successful bond formation and a faster reaction.

It's important to identify the type of chemical reaction such as decomposition, synthesis, single-replacement, or double-replacement, to predict the products formed. Knowing these details enables chemists to manipulate conditions to achieve desired reactions efficiently.
moles and molar mass
The mole is a unit that measures the amount of substance. It is one of the foundational concepts in chemistry, essential for quantifying entities at the molecular and atomic scale. One mole of any substance contains Avogadro's number of particles, approximately \(6.022 \times 10^{23}\) entities, whether they be atoms, molecules, or ions.

Molar mass is the mass of one mole of a substance. It is usually expressed in grams per mole (g/mol). For example, the molar mass of carbon dioxide (\(\mathrm{CO}_2\)) is approximately 44.01 g/mol, calculated by adding the atomic masses of carbon (around 12.01 g/mol) and oxygen (about 16.00 g/mol each for two oxygen atoms).
  • To find the number of moles from the mass, use the formula \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \).
  • Conversely, to find mass from moles, rearrange the formula to \( \text{Mass} = \text{Moles} \times \text{Molar Mass} \).
Understanding moles and molar mass allows chemists to determine reactant quantities needed or product amounts formed in chemical reactions, crucial for both theoretical calculations and practical laboratory work.
chemical equations
Chemical equations represent chemical reactions using symbols and formulas. They provide a detailed accounting of the reactants and products involved. For balancing chemical equations, it is essential because mass and atoms are conserved in a reaction. This means the same number of each type of atom must appear on both sides of the equation.

In the given reaction of sodium bicarbonate (\(\mathrm{NaHCO}_3\)), the balanced chemical equation is:\(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)This equation indicates that two molecules of \(\mathrm{NaHCO}_3\) decompose to form one molecule of \(\mathrm{Na}_2\mathrm{CO}_3\), one molecule of \(\mathrm{CO}_2\), and one molecule of \(\mathrm{H}_2\mathrm{O}\).
  • Coefficients in a chemical equation tell us the proportion of molecules involved in the reaction.
  • Reactants are listed on the left, while products are listed on the right.
  • It's crucial to correctly balance an equation to respect the law of conservation of mass.
By correctly interpreting and using chemical equations, chemists can predict how substances will react, determine yield, and ensure safety in processes.

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

Identify the ions that exist in each aqueous solution, and specify the concentration of each ion. (a) \(0.25 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) (b) \(0.123 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (c) \(0.056 \mathrm{M} \mathrm{HNO}_{3}\)

The following questions may use concepts from this and previous chapters. Two beakers sit on a balance; the total mass is \(167.170 \mathrm{g} .\) One beaker contains a solution of \(\mathrm{KI}\) the other contains a solution of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} .\) When the solution in one beaker is poured completely into the other, the following reaction occurs: $$ 2 \mathrm{KI}(\mathrm{aq})+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{KNO}_{3}(\mathrm{aq})+\mathrm{PbI}_{2}(\mathrm{s}) $$ (IMAGE CANNOT COPY) What is the total mass of the beakers and solutions after reaction? Explain completely.

What volume of \(2.06 \mathrm{M} \mathrm{KMnO}_{4},\) in liters, contains \(322 \mathrm{g}\) of solute?

Spectrophotometry A solution of a dye was analyzed by spectrophotometry, and the following calibration data were collected. $$\begin{array}{cc} \text { Dye Concentration } & \text { Absorbance }(A) \text { at } 475 \mathrm{nm} \\ \hline 0.50 \times 10^{-6} \mathrm{M} & 0.24 \\ 1.5 \times 10^{-6} \mathrm{M} & 0.36 \\ 2.5 \times 10^{-6} \mathrm{M} & 0.44 \\ 3.5 \times 10^{-6} \mathrm{M} & 0.59 \\ 4.5 \times 10^{-6} \mathrm{M} & 0.70 \\ \hline \end{array}$$ (a) Construct a calibration plot, and determine the slope and intercept. (b) What is the dye concentration in a solution with \(A=0.52 ?\)

A Suppose you have \(100.00 \mathrm{mL}\) of a solution of a dye and transfer \(2.00 \mathrm{mL}\) of the solution to a \(100.00-\mathrm{mL}\) volumetric flask. After adding water to the \(100.00 \mathrm{mL}\) mark, you take 5.00 mL. of that solution and again dilute to \(100.00 \mathrm{mL}\). If you find the dye concentration in the final diluted sample is \(0.000158 \mathrm{M},\) what was the dye concentration in the original solution?

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.