If 186.0 liters of gaseous O_{2} at 1.09 atm and 577.0 ºC are required for the model rocket to climb 1000 feet, how many grams of solid KClO

Question

If 186.0 liters of gaseous O_{2} at 1.09 atm and 577.0 ºC are required for the model rocket to climb 1000 feet, how many grams of solid KClO_{3} must be in the rocket engine?

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Vodka 3 years 2021-08-25T08:29:08+00:00 1 Answers 26 views 0

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  1. minhkhoi
    0
    2021-08-25T08:31:03+00:00
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    Answer:

    The right answer is “236.53 g“.

    Explanation:

    The given values are:

    P = 1.09 atm

    V = 186 liters

    The reaction will be:

    ⇒  KClO_3 (s)\rightarrow 2Kcl(s)+3O_2(g)

    The moles of O₂ will be:

    =  \frac{PV}{RT}

    On substituting the values, we get

    =  \frac{1.09\times 186}{0.0821\times 850}

    =  \frac{202.47}{69.785}

    =  2.90 \ moles

    Now,

    1 mole O₂ is produced from

    = \frac{2}{3} \ mol \ KClO_3

    then,

    2.90 mole O₂ is produced from 2 mol KClO₃

    = \frac{2}{3}\times 2.90

    = 1.93 \ mol \ KClO_3

    hence,

    The number of grans of solid in the engine will be:

    = 1.93\times 122.55

    = 236.53 \ g

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