CHEMISTRY | MOLE CONCEPT | Question 29


Lead nitrate decomposes on heating as shown in the equation below.
Pb (NO3)2 heat → 2PbO(S) + 4N02(S) + O2(g)

112dm3 of oxygen gas were collected at STP when a sample of lead nitrate was completely decomposed by heating. Calculate;
a. The mass of the lead nitrate sample.
b. The mass of lead (II) oxide produced.
c. The volume of nitrogen dioxide gas produced at STP.
(Pb = 207, Na =14, 0 = 16; molar volume of a gas at STP = 22.4 dm3)


  • DATA


    Volume of O2 = 112dm3
    Molar volume at STP = 22.4 dm3
    Asked
    a. Mass of Pb(NO3)2
    b. Mass of PbO
    c. Volume of NO2
    d. Equation for the reaction

    2Pb (NO3)2(s) → 2PbO(s) + 4NO(g) + O2(g)

    To convert 112dm3 of O2 into mole
    Mole = volume / Molar volume at STP
    = 5 mole of O2
  • A. To find mass of Pb(NO3)2 from mole equivalent


    2 mole of Pb (NO3)2 Ξ 1 mole O2
    ?? Ξ 5 moles O2

    = (2×5) / 1
    = 10 moles of Pb(NO3)2

    Mass = mole × molar mass
    = 10 moles × 331g/mole
    = 3310g

    3310g of lead nitrate Pb(NO3)2 was used

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