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Sagot :
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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