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24. A laboratory chemist wants to produce 25.0 g NO2 by the following reactions. If the
percent yield on both steps is 82%, how much N20s must he start with?
N205 2 NO +
02
2 NO + 02 → NO2

Sagot :

Oseniidn

The amount of N2O5 to start with would be 35.79 grams

Stoichiometric calculations

From the balanced equation of the reactions:

[tex]2N_2O_5 --- > 4NO_2 + O_2[/tex]

Mole ratio of N2O5 and NO2 = 1:2

Since the reaction's actual yield is 82%, the theoretical yield would be: 25 x 100/82 = 30.49 grams

Mole of 25 g NO2 = 30.49/46

                              = 0.66 mole

Equivalent mole of N2O5 = 0.54/2 mole

                                               = 0.33 moles

Mass of 0.27 mole N2O5 = 0.33 x 108.01

                                                = 35.79 grams

More on stoichiometric calculations can be found here: https://brainly.com/question/8062886

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