Number of moles of $K_{2} P t C l_{6}$ consumed is :

0.048

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0.024

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0.012

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0.096

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Solution

The correct option is B $0.024$
$10 g o f K_{2} P t C l_{6} = \frac{10}{417} m o l = 0.024 m o l$ (limiting)
$10 g N H_{3} = \frac{10}{17} = 0.588 m o l$ (excess)
$K_{2} P t C l_{6} c o n s u m e d = 0.024 m o l$
$N H_{3} c o n s u m e d = 0.048 m o l$
$N H_{3} u n r e a c t e d = 0.588 - 0.048$ $= 0.54 m o l$
$N H_{3}$ is limiting reagent, so $K_{2} P t C l_{6}$ consumed is $0.024$ mol.

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Similar questions

A + B \to C

A + B \to C

\to

\begin{matrix} E x p . & [A]_{0} & [B]_{0} & I n i t i a l r a t e 1 & 0.012 & 0.035 & 0.10 2 & 0.024 & 0.035 & 0.80 3 & 0.012 & 0.070 & 0.10 4 & 0.024 & 0.070 & 0.80 \end{matrix}

A + B \to P r o d u c t s

A + B ⟶ C

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