Answered

A mixture of N2(g) and H2(g) reacts in a closed container to form ammonia, NH3(g). The reaction ceases before either reactant has been totally consumed. At this stage 1.0 mol N2, 1.0 mol H2, and 1.0 mol NH3 are present.
How many moles of N2 and H2 were present originally?



Answer :

Oseni

The moles of [tex]N_2[/tex] and [tex]H_2[/tex] that were present originally in the reaction will be 1.5 moles and 2.5 moles respectively.

Mole ratios in reactions

The two reactants,  [tex]N_2[/tex] and [tex]H_2[/tex], react in a closed container to form [tex]NH_3[/tex]. The normal stoichiometric ratio of the reaction is according to the following equation:

[tex]N_2 (g) + 3H_2 (g) --- > 2NH_3 (g)[/tex]

The stoichiometric ratio of  [tex]N_2[/tex], [tex]H_2[/tex], and [tex]NH_3[/tex] is 1:3:2.

This means for every 1 mole of ammonia formed, 0.5 moles of [tex]N_2[/tex] and 1.5 moles of [tex]H_2[/tex] is required.

However, 1 mole of [tex]N_2[/tex] and 1 moles of [tex]H_2[/tex] remained after the reaction has ceased.

Thus, total amount of  [tex]N_2[/tex] present originally = 1 + 0.5

                                                                           = 1.5 moles

Total amount of  [tex]H_2[/tex] present originally = 1 + 1.5

                                                               = 2.5 moles

More on mole ratios of reactions can be found here: https://brainly.com/question/5441060

#SPJ1

Other Questions