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Ierofanga [76]
3 years ago
6

A 1.018 g sample pure platinum metal was reacted with HCl to form 1.778 g of a compound containing only platinum and chlorine. D

etermine the empirical formula of this "Pt-Cl" Compound.
Chemistry
1 answer:
Ivanshal [37]3 years ago
3 0

Answer:

PtCl_4

Explanation:

Hello!

In this case, since HCl and Pt react according to the following chemical equation:

HCl+Pt\rightarrow PtCl_x+H_2

Whereas PtClx is the compound containing Pt and Cl; thus, since 1.018 g out of 1.778 g correspond to Pt and therefore 0.760 g to chlorine, so we determine the empirical formula of this compound by firstly computing the moles of each element:

n_{Pt}=1.018gPt*\frac{1molPt}{195.084gPt}=0.00522molPt\\\\\\n_{Cl}=0.760gCl*\frac{1molCl}{35.45gCl}  =0.0214molCl

Now, we divide the each moles by those of Pt as the fewest ones in order to compute their subscripts in the empirical formula:

Pt=\frac{0.00522}{0.00522}=1 \\\\Cl=\frac{0.0214}{0.00522} =4

Thus, the required formula is:

PtCl_4

Best regards!

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