Which reactant is limiting and which is in excess?

Limiting -

K_{2} P t C l_{4}

, Excess -

N H_{3}

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Limiting -

N H_{3}

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Exact molar ratio

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Limiting - would he decided by the quantity of the product formed

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Solution

The correct option is A Limiting - $K_{2} P t C l_{4}$ , Excess - $N H_{3}$
$10 g o f K_{2} P t C l_{6} = \frac{10}{417} m o l = 0.024 m o l$ (limiting)
$10 g N H_{3} = \frac{10}{17} = 0.588 m o l$ (excess)
$K_{2} P t C l_{6} c o n s u m e d = 0.024 m o l$
$N H_{3} c o n s u m e d = 0 : 048 m o l$
$N H_{3} u n r e a c t e d = 0.588 - 0.048$ $= 0.54 m o l$
Therefore, the limiting reagent is $K_{2} P t C l_{4}$ and in excess is $N H_{3}$ .

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