Question

N2O5 reacts with water to produce nitric acid. If 1.93 moles of N2O5 react with excess water, how many moles of nitric acid can be produced?

Answer 1

2. Balanced equation is N2O5 + H2O -------> 2 HNO3 ........... 1 mole of N2O5 produces 2 moles of HNO3 .....therefore 1.93 moles of N2O5 will produce 1.93 x 2 = 3.86 moles of HNO3 or nitric acid .............

Answer 2

The balanced chemical equation showing the reaction between dinitrogen pentoxide (N₂O₅) with water H₂O can be computed as follows:

$\mathbf{N_2O_5 + H_2O \to 2HNO_3}$

From the given information, the amount of water H₂O is present in excess, as such dinitrogen pentoxide (N₂O₅) is regarded to be the limiting reagent.

However,

• if 1 mole of dinitrogen pentoxide (N₂O₅) result in 2 moles of HNO₃

∴

1.93 moles of N₂O₅ will result in = (2 × 1.93) moles of HNO₃

= 3.86 moles of HNO₃

Therefore, we can conclude that the reaction between 1.93 moles of N₂O₅ with water will produce 3.86 moles of nitric acid HNO₃.

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