$2 K M n O_{4} + 5 H_{2} C_{2} O_{4} + 3 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 10 C O_{2} + 8 H_{2} O$
How many moles of $M n S O_{4}$ are produced when $1.5$ mol of $K M n O_{4}, 3.5 m o l$ of $H_{2} C_{2} O_{4}$ , and $2.5$ mol of $H_{2} S O_{4}$ are mixed :

0.75

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1.4

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1.5

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1.7

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Solution

The correct option is A $1.4$
$K M n O_{4} + 5 H_{2} C_{2} O_{4} + 3 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 10 C O_{2} + 8 H_{2} O$
$K M n O_{4} = \frac{moles}{coefficient} = 0.75$
$H_{2} C_{2} O_{4} = \frac{moles}{coefficient} = 0.7$
$H_{2} S O_{4} = \frac{moles}{coefficient} = 0.83$
So, $H_{2} S O_{4}$ is the limiting reagent.
So, $5 m o l e s$ produce $2 m o l e s$ .
So, $3 m o l e s$ produce $1.4 m o l e s$ of $M n S O_{4}$ .

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