a 0.2000 grams sample of aluminum metal is added to a solution containing 1.0000grams of an unknown compound made of platinum and chloride ion. the reaction results in the formation of aluminum chloride and platinum metal. when the reaction is complete, unreacted aluminum remains and this unreacted aluminum is consumed by reaction with hydrochloric acid. after washing and drying the, the mass of platinum metal is recovered is 0.5780.
what is the balanced equation for the reaction of aluminum with the original unknown compound?
what is the minimum mass of the aluminum metal needed to react with the platinum chloride compound?
2 answers
also, what is the experimental percent of platinum in the original compound?
The problem doesn't say that the 1.00 g sample is pure; I'll assume it is.
mols Pt = 0.5780/195 = about 0.00296 but you need to do ALL of these calculations more carefully.
g Cl = 1.00-0.5760 = 0.422g and convert to mols = approx 0.012
So the empirical formula is PtCl4.
Equation is
4Al + 3PtCl4 ==> 4AlCl3 + 3Pt
mols Pt = 0.00296
mols Al = 0.00296 x (4 mola Al/3 mols Pt) = about 0.004 mols Al and that x atomic mass Al gives about 0.1 g Al as the minimum mass. Again you should redo all of this.
%Pt (if the compound is a pure compound) is (0.5780/1.00)*100 = ?
mols Pt = 0.5780/195 = about 0.00296 but you need to do ALL of these calculations more carefully.
g Cl = 1.00-0.5760 = 0.422g and convert to mols = approx 0.012
So the empirical formula is PtCl4.
Equation is
4Al + 3PtCl4 ==> 4AlCl3 + 3Pt
mols Pt = 0.00296
mols Al = 0.00296 x (4 mola Al/3 mols Pt) = about 0.004 mols Al and that x atomic mass Al gives about 0.1 g Al as the minimum mass. Again you should redo all of this.
%Pt (if the compound is a pure compound) is (0.5780/1.00)*100 = ?
2 answers