A mixture of calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) and magnesium carbonate \(\left(\mathrm{MgCO}_{3}\right)\) of mass \(6.26 \mathrm{~g}\) reacts completely with hydrochloric acid (HCl) to generate \(1.73 \mathrm{~L}\) of \(\mathrm{CO}_{2}\) at \(48^{\circ} \mathrm{C}\) and 1.12 atm. Calculate the mass percentages of \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgCO}_{3}\) in the mixture.

Short Answer

Expert verified
The mass percentage of \(\mathrm{CaCO}_{3}\) in the mixture is 63.1%, and the mass percentage of \(\mathrm{MgCO}_{3}\) in the mixture is 36.9%.

Step by step solution

01

Identify the reactions

The reactions that take place are: \(\mathrm{CaCO}_{3}(s) + 2\mathrm{HCl}(aq) → \mathrm{CaCl}_{2}(aq) + \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\) \(\mathrm{MgCO}_{3}(s) + 2\mathrm{HCl}(aq) → \mathrm{MgCl}_{2}(aq) + \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\) In these reactions, 1 mole of each type of carbonate gives 1 mole of CO2. This fact will be important when accounting for gas produced.
02

Determine moles of CO2 produced

Use the ideal gas equation: \(PV=nRT\) changing given conditions to required units: \(P=1.12\,atm\), \(V=1.73\,L\), \(T=48+273.15=321.15\,K\) and \(R=0.0821\,atm.L/mol.K\), solve for \(n=(PV)/(RT)=((1.12*1.73)/(0.0821*321.15))=0.073\,moles\). There are 0.073 moles of CO2 produced.
03

Calculate theoretical moles of CO2 each compound could produce

The potential moles of CO2 each compound can produce is equal to the moles of the compound present. Hence, the moles of CaCO3 = moles of CO2 it can produce and moles of MgCO3 = moles of CO2 it can produce. Let's denote the mass (in g) of CaCO3 as \(x\) and of MgCO3 as \(6.26-x\). Hence, the moles of CaCO3 present = \(x/100.09\) and the moles of MgCO3 present = \((6.26-x)/84.31\), as the molar mass of CaCO3 is 100.09 g/mol and that of MgCO3 is 84.31 g/mol. Thus, the amount of CO2 that each compound can produce can be expressed in terms of \(x\).
04

Equate total CO2 to CO2 produced by individual compounds

Since given total moles = sum of moles from each compound, we can write: 0.073 \(moles\) = \(x/100.09 + (6.26-x)/84.31\), which allows us to solve for \(x = 3.95\,g\).
05

Calculate mass percentages

Percent mass of calcium carbonate = \((x/6.26)*100 = 63.1%\). Percent mass of magnesium carbonate = \(100 - 63.1% = 36.9%\)

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 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

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.

Sign-up for free