Asked by Sofia
A 0.509 g sample pure platinum metal was reacted with hcl to form 0.889 g of a compound containing only platinum and chlorine?
Answers
Answered by
bobpursley
So the product must be of the form PtCl<sub>n</sub>, and the question is what is n, the ratio of the chloride to platinum.
Moles of Pt: .509/atomicmassPt =.509/198=.00257
moles of Cl: (.889-.509)/atomicmassCl=.38/35.45=.0107
dividing each by the lowest, to get the ratio
Cl: .107/.00257=4.2
Pt:; .00257/.00257=1
Empirical formula: PtCl<sub>4</sub>
Moles of Pt: .509/atomicmassPt =.509/198=.00257
moles of Cl: (.889-.509)/atomicmassCl=.38/35.45=.0107
dividing each by the lowest, to get the ratio
Cl: .107/.00257=4.2
Pt:; .00257/.00257=1
Empirical formula: PtCl<sub>4</sub>
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.