Chapter 4
Q. 4.15
Acid rain in the form of aqueous H_{ 2}SO_{4} slowly erodes marble statues, as shown in Figure 4.15. If we add 16.75 mL of 0.0100 M H_{ 2}SO_{ 4} to 25.00 mg of CaCO_{ 3}, will all of the CaCO_{ 3} react?
StepbyStep
Verified Solution
Collect and Organize Given the volume and molarity of the H_{ 2}SO_{ 4} solution, we can calculate the mass of CaCO_{ 3} that reacts.
Analyze We need the balanced chemical equation for the reaction to determine the stoichiometry of the reaction between CaCO_{ 3} and H_{ 2}SO_{ 4} . The molecular equation is
CaCO_{ 3}(s)+H_{ 2}SO_{ 4}(aq) →CaSO_{ 4}(s)+H_{ 2}O (\ell)+CO_{2}(g)
One mole of H_{ 2}SO_{ 4} (aq) reacts with 1 mole of CaCO_{ 3} (s). The steps involved in calculating the moles of H_{ 2}SO_{ 4} available and the mass (in milligrams) of calcium carbonate that will react with it are
Solve We calculate the mass of calcium carbonate that reacts, starting with the volume of sulfuric acid available and its molarity:
16.75 \sout{mL H_{ 2}SO_{ 4}}\times \frac{1 \sout{L}}{10^{3} \sout{mL}} \times \frac{1.00\times 10^{2} \sout{mol H_{ 2}SO_{ 4}}}{1 \sout{L H_{ 2}SO_{ 4}}} \times \frac{1 \sout{mol CaCO_{3}}}{1 \sout{mol H_{2}SO_{4}}} \times \frac{100.09 \sout{g CaCO_{3}}}{1 \sout{mol CaCO_{3}}} \times \frac{1000 mg CaCO_{3}}{1 \sout{g CaCO_{3}}} =16.8 mg CaCO_{3}
Only 16.8 mg of the 25.00 mg sample will react to form solid calcium sulfate (Table 4.4).
Table 4.4 Solubility Rules for Common Ionic Compounds in Water 
All compounds containing the following ions are soluble:

Compounds containing the following anions are soluble except as noted:

Insoluble compounds include the following:

Think About It The sulfuric acid is the limiting reactant, since only 67% of the CaCO_{ 3} reacts under the conditions in this exercise.