If .869 grams of O3 reacts with .665 grams of NO, how many grams of NO2 will be produced?

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Is ozone or nitric oxide the limiting reagent?

At the end of the reaction, what is the number of moles of the excess reagent?

Thank you for your time.

1 Answer

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  • equation: O3 + 3NO –> 3NO2

    .869g O3 * 1mole/48g =0.0181 moles, which would produce .0181*3=.0543moles NO2

    .665 grams NO * 1mole/30g=0.0222moles, which would produce the same number of moles of NO2.

    The amount of NO available would produce less product, and is the limiting reagent, and the grams of NO2 produced can be calculated from the 0.0222 moles of NO2 that will be produced from the NO.

    0.0222 moles * 46g/1mole=1.0197g NO2

    For the number of moles of excess reagent left over, the reaction consumed all 0.222 moles of NO, and the moles of NO:moles of O3 ration from the equation is 3:1, so 0.0222/3=0.739 moles of O3 will be consumed, which leaves 0.0181-0.739=0.0107 moles of ozone.

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