Which reactant is the limiting reagent and which is in excess?

K_{2} [P t C l_{4}]

and

N H_{3}

respectively

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N H_{3}

and

K_{2} [P t C l_{4}]

respectively

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Both A and B

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None of the above

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Solution

The correct option is B $K_{2} [P t C l_{4}]$ and $N H_{3}$ respectively
Molecular weight of $K_{2} [P t C l_{4}] = 2 \times 39 + 195 + 4 \times 35.5 = 415$ g/mol
Molecular weight of $N H_{3} = 17$ g/mol
Moles of $K_{2} [P t C l_{4}] = \frac{83.0}{415} = 0.2$ mol (limiting reagent)
Moles of $N H_{3} = \frac{83}{17} = 4.88$ mol (excess)

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