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Basic solutions of $\mathrm{Na}_{4} \mathrm{XeO}_{6}$ are powerful oxidants. What mass of $\mathrm{Mn}\left(\mathrm{NO}_{3}\right) 2 \cdot 6 \mathrm{H}_{2} \mathrm{O}$ reacts with 125.0 $\mathrm{mL}$ of a 0.1717 M basic solution of $\mathrm{Na}_{4} \mathrm{XeO}_{6}$ that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate?

9.8814 $\mathrm{g}$

Chemistry 101

Chapter 18

Representative Metals, Metalloids, and Nonmetals

Nonmetals Chemistry

University of Maryland - University College

University of Kentucky

University of Toronto

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question number 116 is a stoke geometry solution. Stoy, geometry calculation. The hardest part of this problem is simply figuring out what the chemical reaction is and then balancing it. If we try to limit everything to just the ions involved in the chemical reaction, we'll strip off. The sodium here will strip off the sodium here will strip off the nitrates in the water here, and these will be our reactant. They tell us that we have to products Zen on and permanganate sodium permanganate, but we'll assume that sodium is a spectator ion. But in this way we could balance except for this hydrogen right here. So we'll assume that we're also making hydrogen gas, which would have been helpful had they told us that. Then we can balance it by putting twos in the appropriate locations. We now see the stoy geometry between this and the manganese compound is 2 to 1. So if we start with the mill leaders of our compound that's reacting. That's a strong oxidizer. We can then convert it to leaders by dividing by 1000 and then two moles by multiplying by the molar ity of that solution. We then know our stoy. Geometry is 1 to 2, so we can convert moles of this Z E or X E 06 to the four minus is actually a minus for every one mole of manganese two plus and then every one mole of manganese two plus corresponds toe. One mole of that man, Janie's exa hydrate manganese to nitrate exa hydrate. We can then convert the moles manganese to a hex, a hydrate, 2 g manganese to hex hydrate by multiplying by its smaller mass. And that gives us 12.3 g of manganese to hex. A hydrate can be reduced. I'm sorry. Be oxidized. With 125 mL of 1250.1717 Moeller of the Xenon compound solution.

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