Stoichiometry And Reactions

Question

Small quantities of hydrogen gas can be prepared in the laboratory by placing zinc metal in aqueous hydrochloric acid. Typically the hydrogen gas is bubbled through water for ease of collection and it becomes saturated with water vapor. Suppose 240ml of hydrogen gas is collected over water at 30C and a barometric pressure of 1.032 atm. (The vapor pressure of water at 30 o is 32 torr)

a. What is the partial pressure of the hydrogen gas? 

b. How many grams of zinc were reacted to generate this volume of hydrogen?

c. What is the minimum volume of a 50 % by mass solution of hydrochloric acid required to react with all of the zinc? Assume the density of the hydrochloric acid solution is 1.00 g/ml.

d. To ensure all the zinc reacts, 25 % more hydrochloric acid is used than the minimum required. What is the concentration of the ions in solution after the reaction is complete?

Answer 1

Zn(s) + 2HCl(aq) ==> H2(g) + ZnCl2(aq) ... BALANCED EQUATION FOR THE REACTION

a). Total pressure (P_T) = sum of partial pressures (P_H2) + (P_H2O)

P_H2 = P_T - P_H2O

P_H2 = 1.032 - 32/760 = 1.032 - 0.04211 (converted 32 torr to atm by dividing by 760)

P_H2 = 0.990 atm

b). Use ideal gas law to find mols H2 generated. Then use stoichiometry of balanced equation to find moles and mass of Zn used.

PV = nRT

n = PV / RT = (0.990 atm)(0.240 L) / (0.0821 Latm/Kmol)(303K)

n = 0.00955 mols H2

0.00955 mols H2 x 1 mol Zn / 2 mols H2 x 65.4 g Zn / mol = 0.312 g Zn reacted

c). 0.312 g Zn x 1 mol Zn / 65.4 g x 2 mol HCl/mol Zn = 0.00955 mols HCl x 36.5 g / mol = 0.349 g HCl

If density = 1.00 g / ml, then you'd need 0.349 mls HCl if pure (100%). At 50% by mass, you need

0.349 g / 0.5 = 0.698 mls HCl

d). 25% x 0.698 mls = 0.1745 mls extra HCl (excess) = 0.1745 g HCl in excess

0.1745 g HCl x 1 mol / 36.5 g = 0.00478 mol HCl in excess

Assuming the change in volume of the reaction is negligible compared to the 240 mls, the final volume of the reaction is still 0.240 L. To find the concentration of ions in solution, simply divide mols of HCl and mols ZnCl₂ by 0.240 L to get molar concentrations.

Add Cl^- from HCl to that from ZnCl₂ remembering that each M ZnCl₂ is equal to 2 M Cl^-

[H⁺] = [H₃O⁺] will be mols H₃O⁺ / 0.240 L