CHEMISTRY | MOLE CONCEPT | Question 14


Take the reaction
NH3 + O2 → NO + H2

In an experiment 325g of NH3 are allowed to read with 3.5g of O2 a) Which reactant is the limiting reagent?
b) How many gram of NO are formed?
c) How much of the excess reactant remain after the reactant?


  • DATA GIVEN:


    Data
    Mass of NH3 = 3.25g
    Mass of O2 = 3.5g
    Equation
    4NH3(g) + 502(g) → 4NO(g) + 6H20(g)
  • a/ The limiting reagent


    Mole of NH3 = mass /Molar mass
    = 3.25g /17g/mole
    = 0.1911764 mole
    Mole ratio for NH3 = Mole /Coefficient mole
    = 0.1911764 mole /4 mole
    = 0.047794
    Mole for O2 = Mass /Molar mass
    = 3.5g /32g/mole
    = 0.109375
    Mole ratio for O2 = Mole /Mole coefficient
    = 0.109375 /5
    = 0.021875
    a/ The limiting reagent is oxygen gas (O2)
  • b/ To calculate the mass of NO formed


    To use limiting reagent to know the ammoniaformed from mole equivalent
    5 moles of O2 Ξ 4 moles of NO
    0.109375 mole of O2 Ξ ?? of NO

    = 0.109375 × 4 /5
    = 0.0875 moles
    To convert 0.0875 moles of NO into mass
    Mass = 0.0875 moles
    = 2.625g
    2.625g of NO was formed.
  • c/ To find the mass of excess reaction remain from mole equivalent between NH3 and O2


    4 moles of NH3 Ξ 5 mole O2
    ?? Ξ 0.109375 O2

    4 moles × 0.109375 mole /5 moles
    = 0.0875 moles of NH3
    To convert mole into mass.
    Mass = moles × molar mass
    = 0.0875 mole × 17g/mole
    = 1.4875g
    Mass remain of NH3 = original mass - mass reacted
    = 3.25g - 1.4875g
    = 1.7625g
    1.7625g of ammonia (NH3) was remained after reaction

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