`50.0kg` of `N_(2)` `(g)` and `10.0kg` of `H_(2)` `(g)` are mixed to produce `NH_(3)` `(g)`. Calculate the `NH_(3)` `(g)` formed. Identify the limiting reagent in the production of `NH_(3)` in this situation.

50.0 kg of N_(2) (g) and 10 kg of H_(2) (g) are mixed to produce NH_(3) (g) . Calculate the NH_(3) (g) formed. Identify the limiting reagent.

For a reaction, N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g) , identify dihydrogen (H_(2)) as a limiting reagent in the following reaction mixtures.

For a reaction, N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g) , identify dihydrogen (H_(2)) as a limiting reagent in the following reaction mixtures.

Taking one mole of N_(2(g)) and excess of H_(2)(g) , under suitable conditions, formation of NH_(3) is completed when NH_(3)(g) has mole fraction of 0.5 , The initial moles of H_(2) are?

6 g of H_(2) reacts with 14 g N_(2) to form NH_(3) till the reaction completely consumes the limiting reagent. The mass of other reactant (in g) left are ……

28 g of N_(2) and 6 g of H_(2) were mixed. At equilibrium 17 g NH_(3) was produced. The weight of N_(2) and H_(2) at equilibrium are respectively

One mole of N_(2) (g) is mixed with 2 moles of H_(2)(g) in a 4 litre vessel If 50% of N_(2) (g) is converted to NH_(3) (g) by the following reaction : N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) What will the value of K_(c) for the following equilibrium ? NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)

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

What volume of H_(2)(g) is produced by decomposition of 2.4 L NH_(3) (g) ? .

One "mole" of NH_(4)HS(s) was allowed to decompose in a 1-L container at 200^(@)C . It decomposes reversibly to NH_(3)(g) and H_(2)S(g). NH_(3)(g) further undergoes decomposition to form N_(2)(g) and H_(2)(g) . Finally, when equilibrium was set up, the ratio between the number of moles of NH_(3)(g) and H_(2)(g) was found to be 3 . NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g), K_(c)=8.91xx10^(-2) M^(2) 2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g), K_(c)=3xx10^(-4) M^(2) Answer the following: To attain equilibrium, how much % by weight of folid NH_(4)HS got dissociated?