In a reaction vessel, the following reaction was carried out using 0.250 mol of NH3(l) and 0.100 mol of N2(g).

4NH3(l) + N2(g) ¡ú 3N2H4(l)
Which of the following represents the composition in moles in the vessel when the reaction reaches completion?
¡Ö0 mol NH3, 0.038 mol N2, 0.333 mol N2H4
¡Ö0 mol NH3, 0.038 mol N2, 0.188 mol N2H4
¡Ö0 mol NH3, 0.063 mol N2, 0.188 mol N2H4
0.150 mol NH3, ¡Ö0 mol N2, 0.300 mol N2H4

I can't interpret the symbols but I refer you back to the stoichiometry and limiting reagent problems.

To determine the composition in moles in the vessel when the reaction reaches completion, we need to use the stoichiometry of the reaction to calculate the amounts of NH3, N2, and N2H4.

Given:
Initial amount of NH3 = 0.250 mol
Initial amount of N2 = 0.100 mol

The balanced equation for the reaction is:
4NH3(l) + N2(g) → 3N2H4(l)

From the equation, we can see that the mole ratio between NH3 and N2 is 4:1, and the mole ratio between NH3 and N2H4 is 4:3.

To determine the amount of NH3 remaining after the reaction reaches completion, we can use the mole ratio between NH3 and N2:

0.100 mol N2 × (4 mol NH3 / 1 mol N2) = 0.400 mol NH3

To determine the amount of N2H4 formed after the reaction reaches completion, we can use the mole ratio between NH3 and N2H4:

0.100 mol N2 × (3 mol N2H4 / 1 mol N2) = 0.300 mol N2H4

Therefore, the composition in moles in the vessel when the reaction reaches completion is:
0.400 mol NH3, 0 mol N2, and 0.300 mol N2H4.

Among the given options, none of them match the correct composition.