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How many mole of FeCl(3) can be prepared...

How many mole of `FeCl_(3)` can be prepared by the reaction of `10g` pf `KMnO_(4) 10.07` mole of `FeCl_(2)` and `500mL` of `3M HCl` follwing the reaction:
`5FeCl_(2) + KMnO_(4) + 8HCI rarr 5FeCl_(3) + KCl + MnCl_(2) + 4H_(2)O`

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The correct Answer is:
`0.315 mole`
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