×
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 95
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 95

×

# : Antacids are often used to relieve pain and promote

ISBN: 9780321809247 1

## Solution for problem 95 Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants

Chemistry: A Molecular Approach | 3rd Edition

4 5 1 262 Reviews
16
2
Problem 95

95: Antacids are often used to relieve pain and promote healing in the treatment of mild ulcers. Write balanced net ionic equations for the reactions between the aqueous HCI in the stomach and each of the following substances used in various antacids: (a) AI(OH)3(s), (b) Mg(OH)2(s), (c) MgCO3(s), (d) NaAI(CO3)(OH)2(s), (e) CaCO3(s).

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to write the net ionic equations for the reaction between HCl and the given substances used in various antacids. Step1: Antacids are the substances which neutralizes the effect of an acid in our body. They are composed of bases which undergoes neutralization reactions with acid to produce neutral solution of water and salt. Net ionic equation shows only those chemical species which are taking part in a chemical reaction. while writing the ionic equation, we write out all the soluble compounds as ions and eliminate ions common to both the reactants and products. This will give us the net ionic equation. Step2: Let us write the balanced molecular and net ionic equations for the reaction between HCl and the bases present in the antacids: a) HCl reacts with Al(OH) to produce aluminium chloride and water. The balanced 3 molecular equation for the above reaction is: 3HCl(aq) + Al(OH) (s) ----3-> AlCl (aq) + 3H O 3 2 Using the guidelines of solubility, we can say that Al(OH) is insoluble in water3nd hence will not be dissociated into ions. So, the balanced ionic equation for the above reaction is: + - 3+ - 3H + 3Cl + Al(OH) ---------3-> 3H O + Al + 3Cl2Al(OH) being 3 insoluble cannot be converted into ions] Eliminating common ions on both sides of the equation, we get the net ionic equation as follows: + 3+ 3H + Al(OH) (s) --3--------> 3H O + Al 2 Step3: b) HCl reacts with Mg(OH) to produce magn2ium chloride and water. The balanced molecular equation for the above reaction is: 2HCl(aq) + Mg(OH) (s) -------> MgCl (aq) + 2H O 2 2 2 Using the guidelines of solubility, we can say that Mg(OH) is insoluble in water a2 hence will not be dissociated into ions. So, the balanced ionic equation for the above reaction is: 2H + 2Cl + Mg(OH) ------------> 2H O + Mg + 2Cl [Mg(OH) being+ - 2 2 2 insoluble cannot be converted into ions] Eliminating common ions on both sides of the equation, we get the net ionic equation as follows: + 2+ 2H + Mg(OH) (s) ------2----> 2H O + Mg 2 Step4: c) HCl reacts with MgCO to produce magnesium chloride, carbon dioxide and water. 3 The balanced molecular equation for the above reaction is: 2HCl(aq) + MgCO (s) -------> 3Cl (aq) + CO (g) + H O 2 2 2 Using the guidelines of solubility, we can say that MgCO is insoluble in water and 3 hence will not be dissociated into ions. So, the balanced ionic equation for the above reaction is: + - 2+ - 2H + 2Cl + MgCO ------------3H O + Mg + 2Cl + CO2MgCO being 2 3 insoluble cannot be converted into ions] Eliminating common ions on both sides of the equation, we get the net ionic equation as follows: 2H + MgCO (s) -----3-----> H O + CO + Mg 2 2 2+ Step5: d) HCl reacts with NaAl(CO )(OH) to produc3sodium a2minium chloride, carbon dioxide and water. The balanced molecular equation for the above reaction is: 4HCl(aq) + NaAl(CO )(OH) (s) -------> NaCl(aq) + AlCl (aq) + CO (g) + 3H O 3 2 3 2 2 Using the guidelines of solubility, we can say that NaAl(CO )(OH) is insoluble in 3 2 water and hence will not be dissociated into ions. So, the balanced ionic equation for the above reaction is: + - + - 3+ - 4H + 4Cl + NaAl(CO )(OH) -------3---> 3H 2+ Na + Cl + Al + 3Cl 2CO 2 [NaAl(CO )(OH)3eing inso2ble cannot be converted into ions] Eliminating common ions on both sides of the equation, we get the net ionic equation as follows: 4H + NaAl(CO )(OH) (s) ------------> 3H O + Na + Al + CO + 3+ 3 2 2 2 Step6: e) HCl reacts with CaCO to produce ca3ium chloride, carbon dioxide and water. The balanced molecular equation for the above reaction is: 2HCl(aq) + CaCO (s) ------->3aCl (aq) + CO (g) + H 2 2 2 Using the guidelines of solubility, we can say that CaCO is insoluble in water and 3 hence will not be dissociated into ions. So, the balanced ionic equation for the above reaction is: + - 2+ - 2H + 2Cl + CaCO ------------> 3 O + Ca + 2Cl + CO [C2O being 2 3 insoluble cannot be converted into ions] Eliminating common ions on both sides of the equation, we get the net ionic equation as follows: 2H + CaCO (s) ------------> H O + CO + Ca 2+ 3 2 2 ----------------

Step 2 of 3

Step 3 of 3

## Discover and learn what students are asking

Calculus: Early Transcendental Functions : Iterated Integrals and Area in the Plane
?In Exercises 1 - 10, evaluate the integral. $$\int_{x^{3}}^{\sqrt{x}}\left(x^{2}+3 y^{2}\right) d y$$

#### Related chapters

Unlock Textbook Solution