How many moles of nitric acid will be produced when .51 mole of dinitrogen pentoxide reacts
To determine the number of moles of nitric acid produced when 0.51 moles of dinitrogen pentoxide (N2O5) react, we need to understand the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the reaction between dinitrogen pentoxide and water is:
N2O5 + H2O -> 2HNO3
From the equation, we can see that for every 1 mole of N2O5, 2 moles of HNO3 are produced.
Therefore, if we have 0.51 moles of N2O5, we can calculate the number of moles of HNO3 produced using the stoichiometric ratio:
0.51 moles N2O5 x (2 moles HNO3 / 1 mole N2O5) = 1.02 moles HNO3
So, when 0.51 moles of dinitrogen pentoxide react, 1.02 moles of nitric acid will be produced.
To determine the number of moles of nitric acid produced when a certain amount of dinitrogen pentoxide reacts, we need to use the balanced chemical equation for the reaction.
The balanced equation for the reaction between dinitrogen pentoxide (N2O5) and water (H2O) to produce nitric acid (HNO3) is:
N2O5 + H2O -> 2 HNO3
From this equation, we can see that for every 1 mole of N2O5 that reacts, 2 moles of HNO3 are produced.
Given that we have 0.51 moles of N2O5, we can calculate the moles of HNO3 produced using the stoichiometry of the reaction:
0.51 moles N2O5 * (2 moles HNO3 / 1 mole N2O5) = 1.02 moles HNO3
Therefore, when 0.51 mole of dinitrogen pentoxide reacts, 1.02 moles of nitric acid will be produced.
with what? H2O?
N2O5 + H2O ==> 2HNO3
0.51 mols N2O5 will produce twice that mols HNO3.