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28 September, 15:33

Nitrogen dioxide (NO2) gas and liquid water (H2O) react to form aqueous nitric acid (HNO3) and nitrogen monoxide (NO) gas. Suppose you have 5.0 mol of

NO2 and 11.0 mol of H2O in a reactor.

Suppose as much as possible of the NO2 reacts. How much will be left? Round your answer to the nearest 0.1 mol.

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  1. 28 September, 15:44
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    9.3 moles of H20

    Explanation:

    First, let us write a balanced equation for the reaction. This is illustrated below:

    3NO2 + H2O - > 2HNO3 + NO

    Data obtained from the question include:

    Number of mole of NO2 = 5moles

    Number of mole of H2O = 11moles

    From the balanced equation,

    3 moles of NO2 reacted with 1 mole of H2O.

    Therefore, 5 moles of NO2 will react with = 5/3 = 1.7 mole of H2O

    From the calculation above, it is obvious that some amount of H20 did not react.

    The unreacted H2O = 11 - 1.7 = 9.3 moles

    Therefore, 9.3 moles of H20 is left over
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