If 7.40 g of O3 reacts with 0.670 g of NO, how many grams of NO3 will be produced? Identify the limiting reagent from the reaction.

2O3 + 3NO → 3NO3



O2 produces____grams of NO2



NO produces____grams of NO2

Respuesta :

The limiting reagent is NO and the mass of NO₃ formed in the reaction is 1.38 grams

Limiting reagent:

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and they determine the amount of product formed in a reaction.

2O₃ + 3NO → 3NO₃

From the balance equation 2 moles of trioxide reacted with 3 moles of nitrogen monoxide to produce 3 moles of nitric oxide.

molarity (O₃) = 7.40 / 48 = 0.15416666666 moles

molarity(NO) = 0.670 / 30.01 = 0.02232589136 moles

Therefore,

O₃  :  NO

 2  :  3

Therefore, the limiting reagent is NO. It will completely be used before the reaction comes to an end.

3(30.1) g of NO produces 3(62) g of NO₃

0.670 g of NO will produce ?

cross multiply

mass of NO₃ formed = 0.670 × 186 / 90.3 = 124.62 / 90.3 = 1.38006644518 = 1.38 grams.

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