Question

In: Chemistry

Tums are taken to reduce acid in the stomach. Calcium carbonate (Tums) is reacted with 1.0...

Tums are taken to reduce acid in the stomach. Calcium carbonate (Tums) is reacted with 1.0 M hydrochloric acid in the stomach. Carbon dioxide, water and calcium chloride are formed

a. write  balance chemical equation

b. Write the balanced chemical equations for the formation of each compound from its constituent elements. (5) (Show that they all sum to the balanced chemical equation in part a.)

c. Using the following table of ΔH°f to determine the ΔHrxn. Indicate if the reaction is exothermic or endothermic. (10)

Compound

ΔH°f (kJ/mol)

CaCO3 (s)

−1206.9

HCl (aq)

−167.2

CO2 (g)

−393.509

H2O (l)

−285.8

CaCl2 (aq)

−877.3

d. If you take a 1000 mg tablet, how much energy would be evolved in the reaction with excess hydrochloric acid in your stomach.

Solutions

Expert Solution

a)

HCl + CaCO3 = H2O+ CaCl2

balance

2HCl + CaCO3 = 2H2O+ CaCl2

add phases

2HCl(aq) + CaCO3(s) = 2H2O(L) + CaCl2(aq)

b. Write the balanced chemical equations for the formation of each compound from its constituent elements. (5) (Show that they all sum to the balanced chemical equation in part a.)

Ca(s) + C(s,graphite) + 3/2O2(g) --> CaCO3(s)

1/2H2(g) + 1/2Cl2(g) --> HCl(aq)

H2(g) + 1/2O(g) = H2O(l)

Ca(s) + 2Cl(g) --> CaCl2(aq)

when adding them:

add:

H2(g) + 1/2O(g) = H2O(l)

Ca(s) + 2Cl(g) --> CaCl2(aq)

substract

Ca(s) + C(s,graphite) + 3/2O2(g) --> CaCO3(s)

1/2H2(g) + 1/2Cl2(g) --> HCl(aq)

Overall

2HCl(aq) + CaCO3(s) = 2H2O(l) + CaCl2(aq)

c. Using the following table of ΔH°f to determine the ΔHrxn. Indicate if the reaction is exothermic or endothermic. (10)

HRxn= Hproduct s- HReactatns

HRxn = (2H2O(L) + CaCl2(aq)) - (2HCl(aq) + CaCO3(s))

HRxn = (2*-285.8 + -877.3) - (2*-167.2 + -1206.9)

HRxn = 92.4 kJ/mol

d. If you take a 1000 mg tablet, how much energy would be evolved in the reaction with excess hydrochloric acid in your stomach.

m = 1000 mg tablet

1 g of CaCO3

mol = mass/MW = 1/100 = 0.01 mol of CaCO3

1 mol = 92.4

0.01 = 92.4 *0.01 = 0.924 kJ absorbed


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