Number of moles of cis-platin formed is :

0.012

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0.024

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0.048

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0.192

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Solution

The correct option is B $0.024$
$10 g o f K_{2} P t C l_{6} = \frac{10}{415} m o l = 0.024 m o l$ (limiting)

$10 g N H_{3} = \frac{10}{17} = 0.588 m o l$ (excess)

$K_{2} P t C l_{6} c o n s u m e d = 0.024 m o l = c i s - p l a t i n f o r m e d$

Option B is correct.

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Similar questions

P t (N H_{3})_{2} C l_{2}

c i s - p l a t i n e

\frac{24}{5} = 4.8 = 2.4 \times 2

\frac{24}{25} = 0.96 = 0.24 \times 4

\frac{24}{125} = 0.192 = 0.024 \times 8

A + B \to C

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