What is the maximum amount of $I F_{7}$ $IF_7$ that can be obtained from 25.0 g fluorine in the equation $I_{2} + F_{2} \to I F_{7}$ $I_2 + F_2 -> IF_7$ ?

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What is the maximum amount of $I F_{7}$ $IF_7$ that can be obtained from 25.0 g fluorine in the equation $I_{2} + F_{2} \to I F_{7}$ $I_2 + F_2 -> IF_7$ ?

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Answer 1 · 2016-11-29T05:37:55

You need a stoichiometric equation. You can make a mass of approx. $50 \cdot g$ $50*g$

Explanation:

$\frac{1}{2} I_{2} (g) + \frac{7}{2} F_{2} (g) \to I F_{7}$ $1/2I_2(g) + 7/2F_2(g)rarr IF_7$

I introduced the half-coefficient because it makes the arithmetic a little bit easier. I am certainly free to do so. The stoichiometry dictates that 1 equiv of diiodine reacts with 7 squiv difluorine.

$Moles of difluorine$ $"Moles of difluorine"$ $=$ $=$ $\frac{25.0 \cdot g}{2 \times 19 \cdot g \cdot m o l^{- 1}} = 0.658 \cdot m o l$ $(25.0*g)/(2xx19*g*mol^-1)=0.658*mol$ .

Given stoichiometric iodine, we can form $\frac{2}{7}$ $2/7$ equiv of the interhalogen, i.e. $\frac{2}{7} \times 0.658 \cdot m o l = 0.188 \cdot m o l$ $2/7xx0.658*mol=0.188*mol$ .

And thus a mass of $0.188 \cdot m o l \times 259.90 \cdot g \cdot m o l^{- 1} = ? ?$ $0.188*molxx259.90*g*mol^-1=??$ .

It would not be my choice to do this reaction. I am too fond of my 10 fingers, and 2 eyes.

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What is the maximum amount of $I F_{7}$ $IF_7$ that can be obtained from 25.0 g fluorine in the equation $I_{2} + F_{2} \to I F_{7}$ $I_2 + F_2 -> IF_7$ ?

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Explanation:

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What is the maximum amount of IF7IF_7 that can be obtained from 25.0 g fluorine in the equation I2+F2→IF7I_2 + F_2 -> IF_7?

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What is the maximum amount of $I F_{7}$ $IF_7$ that can be obtained from 25.0 g fluorine in the equation $I_{2} + F_{2} \to I F_{7}$ $I_2 + F_2 -> IF_7$ ?