A method used by the U.S. Environmental Protection Agency
(EPA) f...

Question

A method used by the U.S. Environmental Protection Agency
(EPA) for determining the concentration of ozone in air is to pass
the air sample through a “bubbler” containing sodium iodide,
which removes the ozone according to the following equation:
O31g2 + 2 NaI1aq2 + H2O1l2¡
O21g2 + I21s2 + 2 NaOH1aq2
(a) How many moles of sodium iodide are needed to remove
5.95 * 10-6 mol of O3?

Accepted Answer

Hello everyone today. We have the following problem, laurel sulfate is an anti on founding cleaners and detergents. Its structure is shown below assuming laurel sulfate did composes according to the following reaction here, Calculate the mass of oxygen required to decompose 15 g of laurel sulfate. So the first thing we want to do to find our mass of oxygen gas. Start with what we're given, which is 15 g of laurel sulfate, which has the following chemical formula C 12 H 25 S 04 -. And what we wanna do is we want to take that mass and we want to convert that into moles of laurel sulfate. And so by doing that, we can say that one mole of laurel sulfate Is equal to its molar mass. And according to the periodic table, that's going to be 265. g. And that's of laurel sulfate. Our units for grams of laurel sulfate are going to cross out. Next we want to convert from our moles of laurel sulfate, two moles of oxygen. And by doing that, we can use the multiple ratio that when we have one mole of our laurel sulfate and that one more comes from the fact that this coefficient in front of laurel sulfate is one. We have 18 moles of oxygen gas and that comes from the coefficient of 18 and so our units for moles of laurel sulfate cross out. And so finally, what we can do is multiply that moles of oxygen gas into by its molar mass to find the grams. And so we have one mole of oxygen gas is going to be equal to 32 g of oxygen gas per the periodic table. Each mole of oxygen is worth 16 g, and so we have two of them, So 32 are moles of oxygen gas, cancel out our units, and so we end up with a mass of oxygen gas being 32.6 g as our final answer. And with that we have answered the question. I hope this helps until next time.

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