If a mixture of `3` mol of `H_(2)` and `1 "mole"` of `N_(2)` is completely converted into `NH_(3)`, what would be the ratio of the initial and final volume at same temperature and pressure?

If a mixture of 3 moles of H_(2) and one mole of N_(2) is completely converted into NH_(3) . What would be the ratio of the initial and final volume at same temperature and pressure ?

If a mixture of 3 "mole" of H_(2) and 1 "mole" of N_(2) is completely converted into NH_(3) , what would be the final volume at same P and T ?

If a mixture of 3 mole H_2 and 1 mole of N_2 is completely converted into NH_3 what would be final volume at same P and T.

40% of a mixture of 2.0 mol of N_(2) and 0.6 mol of H_(2) reacts to give NH_(3) according to the equation: N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

40% of a mixture of 0.2 mol of N_(2) and 0.6 mol of H_(2) reacts to give NH_(3) according to the equation: N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

40% of a mixture of 0.2 mole of N_2 and o.6 mole of H_2 react to give NH_3 according to the equation: N_ 2 ( g ) + 3 H_ 2 ( g ) ⇔ 2 N H_ 3 (g) at constant temperature and pressure. Then what is the ratio of the final volume to the initial volume of gases ?

If 340 g of a mixture of N_(2) and H_(2) in the correct ratio gas a 20% yield of NH_(3) . The mass produced would be: