A piece of zinc weighing 15 g was placed in 100 ml of 2 M hydrochloric acid solution.

| July 15, 2021 | education

A piece of zinc weighing 15 g was placed in 100 ml of 2 M hydrochloric acid solution. Determine the molar concentration of salt in the solution after the end of the reaction. Neglect the change in the volume of the solution during the reaction.

Given:
m (Zn) = 15 g V solution (HCl) = 100 ml C (HCl) = 2 M
Find: C (salt) -?
Decision:
1) V solution (HCl) = 100 ml = 0.1 l; C (HCl) = 2 M = 2 mol / L; 2) Write the reaction equation: Zn + 2HCl => ZnCl2 + H2 ↑ 3) Find the amount of substance Zn: n (Zn) = m (Zn) / Mr (Zn) = 15/65 = 0.23 mol; 4) Find the amount of the substance HCl: n (HCl) = C (HCl) * V solution (HCl) = 2 * 0.1 = 0.2 mol; 5) Compare the amount of Zn and HCl: The amount of HCl is in short supply, therefore, calculations will be carried out for it; 6) Find the amount of the substance ZnCl2 (taking into account the reaction equation): n (ZnCl2) = n (HCl) / 2 = 0.2 / 2 = 0.1 mol; 7) Find the molar concentration of ZnCl2 in the solution: Since by condition, neglect the change in the volume of the solution, then V solution (ZnCl2) = V solution (HCl) = 0.1 l; C (ZnCl2) = n (ZnCl2) / V solution = 0.1 / 0.1 = 1 mol / l = 1 M.
Answer: The molar concentration of ZnCl2 is 1 M.



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