Question

how many liters of NH3 will be produced from the reaction of 4.2 L of hydrogen gas with 2.0 L of nitrogen has. Assume that all gases are at the same temperature and pressure. (it's a limiting reagent gas problem!)

balanced equation:
N2(g) + 3H2(g) => 2NH3 (g)

Answer 1

Naig) + 3H2(g) → 2NH3lg) Gives - ھ صململ شرط با 2 Volume of Na= 2.0L 42L 0.1% Smol Moles: of H₂ Volume of_Hz 22.4 L/mol 2204 L/mol Moles - Volume of M2 2001 20.089 3mol. ماه اعلاميه او فهم of from the reaction that I mol of No to produce a mol I so, molee of No = . 1 equation we can conclude reacts with 3 mol of H₂. of NH3 . molee of He = moles of the 3 2 Here so Na But . moles of Ma = 0.0893 to completely least with 0.0893 mol of 0.2679 moles of H2 will be required. 0.1875 mol of Ha is present. He is the limiting reagent. = reacted x2 moles of NH3 produced - = molee of he 3__ :1895 x2 < 3 0.12s'mol Tivolume of Droduced NH3 - molesx 22.4 L = 0.125x2204 = 208 L.

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