Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 127

A 6.53 -g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces 1.72 \(\mathrm{L}\) of carbon dioxide gas at \(28^{\circ} \mathrm{C}\) and 743 torr pressure. (a) Write balanced chemical equations for the reactions that occur between hydrochloric acid and each component of the mixture. (b) Calculate the total number of moles of carbon dioxide that forms from these reactions. (c) Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

Short Answer

Expert verified
The balanced chemical equations for the reactions between hydrochloric acid and magnesium carbonate, and calcium carbonate are: \(MgCO_3 + 2HCl ⟶ MgCl_2 + H_2O + CO_2\) \(CaCO_3 + 2HCl ⟶ CaCl_2 + H_2O + CO_2\) Using the given conditions and the ideal gas law, we find that there are approximately 0.0732 moles of \(CO_2\) produced. Let x be the moles of \(MgCO_3\), so (0.0732 - x) are the moles of \(CaCO_3\). After solving for x, we find that there are approximately 0.0597 moles of \(MgCO_3\), corresponding to a mass of 5.03 g. The percentage by mass of magnesium carbonate in the mixture is approximately 77.1%.

Step by step solution

01

step 1: Balanced chemical equations

For magnesium carbonate and calcium carbonate reacting with hydrochloric acid, the unbalanced reactions are: \(MgCO_3 + HCl ⟶ MgCl_2 + H_2O + CO_2\) \(CaCO_3 + HCl ⟶ CaCl_2 + H_2O + CO_2\) Now we balance the reactions: \(MgCO_3 + 2HCl ⟶ MgCl_2 + H_2O + CO_2\) \(CaCO_3 + 2HCl ⟶ CaCl_2 + H_2O + CO_2\)
02

step 2: Use ideal gas law to find moles of \(CO_2\)

We are given the volume (1.72 L), temperature (28°C), and pressure (743 torr) of the carbon dioxide gas produced. First, we need to convert the temperature to Kelvin: \(T(K) = 28 + 273.15 = 301.15 K\) Next, convert pressure from torr to atm: \(P(atm) = 743 \frac{torr}{760 \frac{torr}{atm}} \approx 0.9776 atm\) Now use the ideal gas law formula to find the moles of \(CO_2\): \(PV=nRT\) where P = pressure (atm), V = volume (L), n = number of moles, R = gas constant = 0.0821 (L atm mol^(-1) K^(-1)), T = temperature (K) \(n = \frac{PV}{RT}\) Solving for n: \(n = \frac{(0.9776)(1.72)}{(0.0821)(301.15)}\) \(n ≈ 0.0732\) moles of \(CO_2\)
03

step 3: Calculate the percentage of MgCO3 in the mixture

Let x be the moles of \(MgCO_3\), so (0.0732 - x) are the moles of \(CaCO_3\). Based on the stoichiometry of the balanced chemical equations, one mole of each carbonate produces one mole of \(CO_2\). Hence, the masses of the carbonates in the mixture are: -Mass of \(MgCO_3 = x × M_{MgCO_3}\), where \(M_{MgCO_3} = 84.31 g/mol\) -Mass of \(CaCO_3 = (0.0732 - x) × M_{CaCO_3}\), where \(M_{CaCO_3} = 100.09 g/mol\) The mass of the mixture is given as 6.53 g. Therefore: \(84.31x + 100.09(0.0732 - x) = 6.53\) Solving this equation for x: \(x ≈ 0.0597\) moles of \(MgCO_3\) Now compute the mass of magnesium carbonate (MgCO3) and then find the percentage of MgCO3 in the mixture. Mass of \(MgCO_3 = 0.0597 × 84.31 = 5.03 g\) Percentage of \(MgCO_3 = \frac{5.03}{6.53} × 100 = 77.1\%\) The percentage by mass of magnesium carbonate in the mixture is approximately 77.1%.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of \(\mathrm{Cl}_{2}\) gas is 8.70 \(\mathrm{L}\) at 895 torr and \(24^{\circ} \mathrm{C}\) .(a) How many grams of \(\mathrm{Cl}_{2}\) are in the sample? (b) What volume will the \(\mathrm{Cl}_{2}\) occupy at \(\mathrm{STP}\) ? (c) At what temperature will the volume be 15.00 \(\mathrm{L}\) if the pressure is \(8.76 \times 10^{2}\) torr? (d) At what pressure will the volume equal 5.00 L if the temperature is \(58^{\circ} \mathrm{C}\) ?

Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at \(80^{\circ} \mathrm{C}\) if 1.00 mol occupies \(33.3 \mathrm{L},\) assuming that (a) \(\mathrm{CCl}_{4}\) obeys the ideal-gas equation; (b) \(\mathrm{CCl}_{4}\) obeys the van der Waals equation. (Values for the van der Waals constants are given in Table \(10.3 .\) ) (c) Which would you expect to deviate more from ideal behavior under these conditions, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4}\) ? Explain.

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-},\) forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-} .\) (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$\mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s)$$ If you allow 15.0 \(\mathrm{g}\) of \(\mathrm{NaClO}_{2}\) to react with 2.00 \(\mathrm{L}\) of chlorine gas at a pressure of 1.50 atm at \(21^{\circ} \mathrm{C},\) how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

A sample of 5.00 \(\mathrm{mL}\) of diethylether \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\) density \(=0.7134 \mathrm{g} / \mathrm{mL}\) ) is introduced into a 6.00 -L vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2},\) whose partial pressures are \(P_{\mathrm{N}_{2}}=0.751 \mathrm{atm}\) and \(P_{\mathrm{O}_{2}}=0.208\) atm. The temperature is held at \(35.0^{\circ} \mathrm{C},\) and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.

The physical fitness of athletes is measured by \(^{u} V_{\mathrm{O}_{2}} \max _{2}^{\prime \prime}\) which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a \(V_{\mathrm{O}_{2}}\) max of 45 \(\mathrm{mL} \mathrm{O}_{2 / \mathrm{kg}}\) body mass/min, but a world-class male athlete can have a \(V_{\mathrm{O}_{2}}\) max reading of 88.0 \(\mathrm{mL} \mathrm{O}_{2} / \mathrm{kg}\) body mass/min. (a) Calculate the volume of oxygen, in mL, consumed in 1 by an average man who weighs 185 lbs and has a \(V_{\mathrm{O}_{2}}\) max reading of 47.5 \(\mathrm{mLO}_{2} / \mathrm{kg}\) body mass/min. (b) If this man lost \(20 \mathrm{lb},\) exercised, and increased his \(V_{\mathrm{O}_{2}}\) max to 65.0 \(\mathrm{mL}\) O \(_{2} / \mathrm{kg}\) body mass/min, how many mL of oxygen would he consume in 1 \(\mathrm{hr}\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation
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.

Sign-up for free