Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

One harmful effect of acid rain is the deterioration of structures and statues made of marble or limestone, both of which are essentially calcium carbonate. The reaction of calcium carbonate with sulfuric acid yields carbon dioxide, water, and calcium sulfate. Because calcium sulfate is marginally soluble in water, part of the object is washed away by the rain. Write a balanced chemical equation for the reaction of sulfuric acid with calcium carbonate.

Short Answer

Expert verified

A balanced chemical equation for the reaction of sulfuric acid with calcium carbonate is,

CaCO3(s)+H2SO4(aq)CaSO4(s)+H2O(l)+CO2(g)

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Definition of Acid rain

Any type of precipitation that contains acidic elements, such as sulfuric or nitric acid, and falls to the earth from the atmosphere in wet or dry forms is referred to as acid rain.

02

Identification of balanced chemical equation

In the troposphere, SO2is almost all oxidized to H2SO4in the form of an aerosol, which ends up in wet precipitation or acid rain.The mechanism for the conversion of SO2to H2SO4is quite complex and not fully understood SO2. can also be oxidized to role="math" localid="1663824190588" SO3 and then converted to H2SO4on particles by heterogeneous catalysis.Eventually, acid rain can corrode limestone and marble CaCO3.

A typical reaction is shown below.

CaCO3(s)+H2SO4(aq)CaSO4(s)+H2O(l)+CO2(g)

Hence, the chemical equation for the reaction of sulfuric acid with calcium carbonate is,

CaCO3(s)+H2SO4(aq)CaSO4(s)+H2O(l)+CO2(g)

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

What is the molecular structure for each of the following molecules or ions?

a. OCl2

b.ClO4-ClO4-

c.ICl5

d.PF6-

What are the Lewis structures for the two allotropic forms of oxygen? How can the Para magnetism of O2be explained using the molecular orbital model? What are the molecular structure and the bond angles in ozone?

The major industrial use of hydrogen is in the production of ammonia by the Haber process:

3H2(g) +N2(g)2NH3(g)

a. Using data from Appendix 4 , calculateH0 ,S0 , and G0for the Haber process reaction.

b. Is this reaction spontaneous at standard conditions?

c. At what temperatures is the reaction spontaneous at standard conditions? Assume thatH0 andS0 do not depend on temperature.

Although nitrogenNF3is a thermally stable compound, nitrogen triiodideNF3is known to be a highly explosive material.NI3can be synthesized according to the equation

BN(s) + 3IF(g)BF3(g) + NI3(g)

  1. What is the enthalpy of formation for NI3(s) given the enthalpy of (-307kJ) and the enthalpies of formation for BN(s)(-254kJ/mol),IF(g)(-96kJ/mol) , and BF3(g)(-1136kJ/mol) ?
  2. It is reported that when the synthesis of NI3is conducted using 4 moles of IF for every 1 mole of BN, one of the by-products isolated is IF2+BF4-


What are the molecular geometries of the species in this by-product? What are the hybridizations of the central atoms in each species in the by-product?

Lead forms compounds in the +2and +4oxidation states. All lead (ll) halides are known (and are known to be ionic). OnlyPbF4 andPbCl4 are known among the possible lead (IV) halides. Presumably lead (IV) oxidizes bromide and iodide ions, producing the lead (II) halide and the free halogen:

PbX4PbX2+X2

Suppose25.00g of a lead (IV) halide reacts to form16.12g of a lead (II) halide and the free halogen. Identify the halogen.

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