A mixture of aluminum and zinc metal is used in 2 experiments described below:
A- A 1.7 gram sample of the mixture is reacted with excess hydrochloric acid and the resulting hydrogen gas is collected over water. The reaction takes place at 750 mm Hg atmospheric pressure and 25C. VP of water = 23.8 torr.
B- In a separate experiment, 1.4 grams of the mixture is reacted with a stoichiometric quantity of hydrochloric acid. The resulting solution is treated with a concentrated ammonia solution. A precipitate forms that is collected and dried. The mass of the precipitate is 2.70 gms.
Calculate the volume of the hydrogen gas collected in A.
Any help would be appreciated, even if it is just where to start
Im mainly confused on how you know what amount of aluminum or zinc is in the samples of the mixture
You find that from part B. The reaction in part B is
Zn + 2HCl ==> ZnCl2 + H2
2Al + 6HCl ==> 2AlCl3 + 3H2
At this point you have Zn^+2 ions and Al^+3 ions in solution. Treatment with concd NH3 solution forms Al(OH)3 with the aluminum ions and Zn(NH3)4^+2 with the zinc ions. So convert 2.70 g of the ppt formed [Al(OH)3] to grams Al and subtract from the total to arrive at the grams Zn. I would get a percent at this point. Then go back to part A, use the percentage to find the amount Zn and Al in the 1.7 g sample, then through the stoichiometry to find the moles H2 from Zn and moles H2 from Al, then use PV = nRT to solve for volume in liters.
Where does the Al(OH)3 come from and how do you know and how do you tell how much of the 2.70g of ppt is Al(OH)3 or Zn(NH3)4^+2??
The Al(OH)3 comes from this reaction. The NH3 + H2O ==> NH4^+ + OH^- provides the OH^- for the reaction.
Al^+3(aq) + 3OH^-(aq) ==> Al(OH)3(s)
The Zn ion forms the zinc ammonia complex ion which is soluble so ALL of the 2.70 g is Al(OH)3. Convert 2.70 g Al(OH)3 to grams Al metal, then
1.4-g Al metal = grams Zn metal.
(Note: The Zn^+2 forms a Zn(OH)2 ppt too; however, excess NH3 dissolves that to form the Zn(NH3)4^+2 (more probably a mixture of Zn(NH3)4^+2 and Zn(OH)4^-2, and both are soluble in excess NH3 especially if concentrated NH3 is used. This is a standard procedure in qualitative analysis but usually one adds NH4Cl to the NH3. In such a case, the Al ppts as Al(OH)3 and the NH4Cl shifts the NH3 equilibrium enough that OH^- is kept low and none of the Zn(OH)2 ppts at all since Ksp for (Zn^+2)(OH^-)^2 is never exceeded due to the low (OH^-).)
Okay, thank you so much! You really know your chemistry
And thank you. But I'm SUPPOSED to know chemistry; therefore, I don't get extra points for that.