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When fluorine gas is passed over solid $M n I_{2}, M n F_{3} and I F_{5}$ are formed. What are the masses of $I F_{5}$ in grams formed by treating $1.54 g of M n I_{2} with 25.0 g of F_{2}$
At wts : Mn = 55, I = 127, F = 19

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answer is 2.21.

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Detailed Solution

$\underset{618}{2 M n I_{2}} + \underset{494}{13 F_{2}} \to \underset{224}{2 M n F_{3}} + \underset{888}{4 I F_{5}}$
618g of $M n I_{2}$ give 888 gr of $I F_{5}$
1.54g of $M n I_{2}$ give = $\frac{888 \times 1.54}{618} = 2.21$

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