If at the beginning of the reaction you used 5.29 g N2 and 9.80 g O2,...

If at the beginning of the reaction you used 5.29 g N₂ and 9.80 g O₂, what is the theoretical yield (in grams) of N₂O₅ (you will need to find the limiting reagent and then the theoretical yield of N₂O₅from this reagent)?

Expert Solution

The balanced reaction is

N2 + 2.5 O2 ------> N2O5

N2 moles = mass of N2 / molar mass of N2 = 5.29 g / 28 g/mol = 0.18893

O2 moles = 9.8 g / 32g/mol = 0.30625

as per reaction O2 moles needed = 2.5 x N2 moles = 2.5 x 0.18893 = 0.47

But we have only 0.30625 moles O2

Hence O2 being relatively less in moles is the limiting reagent

N2O5 moles produced = (1/2.5) x O2 moles = ( 1/2.5) x 0.30625 = 0.1225

N2O5 mass expected = N2O5 moles expected x molar mass of N2O5

= 0.1225 mol x 108 g/mol = 13.23 g

Thus theoretical yield of N2O5 = 13.23 g

queen_honey_blossom answered 2 years ago

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