Determine whether HCl can dissolve each metal sample. If it can, write a balanced chemical reaction showing how the metal dissolves in HCl and determine the minimum volume of 3.25 M HCl required to completely dissolve the sample.(6 pts.) 2.25 g Al 3.76 g Cu 2.48 g Ag

Aluminium can dissolve in HCl

6HCl + 2Al ---> 2AlCl₃ + 3H₂

Minimum volume of HCl required is 76.9 mL.

Explanation:

When a metal dissolves in an acid, a chemical reaction takes place. The chemical reaction occuring is the displacement of the hydrogen ions present in the acid by the metallic ions formed when the metal dissolves in the acid. The displaced hydrogen ion accepts electrons frommthe metal and is evolved as hydrogen gas.

The ability of a metal to be dissolve in an acid with the resultant evolution of hydrogen gas is determined by the position of the metal in the electrochemical series. Metals which are above hydrogen in the electrochemical series are able to displace hydrogen from dilute acids and as such, dissolve in the acid. However, metals which are lower than hydrogen in the electrochemical series are not able to displace hydrogen from dilute acids and as such are not soluble in the acids.

Of the three metals, aluminium, Al, copper, Cu, and silver, Ag, only aluminium is higher than hydrogen in the electrochemical series and as such can dissolve in the 3.25 M HCl.

The equation of the reaction is given below:

6HCl + 2Al ---> 2AlCl₃ + 3H₂

Molar mass of Al = 27 g; Mass of Al reacting = 2.25 g

Number of moles = mass/molar mass

Number of moles of Al present in 2.25 g = 2.25/27 = 0.083 moles

From the equation of reaction, 6 moles of HCL are required to react with 2 moles of Al.

Number of moles of HCl required to react with 0.083 moles of Al = 0.083 × 6/2 = 0.25 moles of HCl

From the formula, molarity = number of moles / volume; volume = number of moles / molarity

Molarity of HCl = 3.25 M; number of molesnof HCl = 0.25 moles

Volume = 0.25 / 3.25 = 0.0769 L or 76.9 mL

Therefore, minimum volume of HCl required is 76.9 mL.