1a)
    i) TABULATE
    burette reading/cm^3 ___ 1 ____ 2 _____ 3
    final reading__________ 39.5 __ 29.80 __ 19.40
    initial reading ______ 19.55 __ 10.50 ___ 0.0
    volume of E used _____ 19.70 ___ 19.30 __ 19.40
    ii) Na2CO3(aq)^2HNO3(aq) — 2NaNO3(aq) + CO2(g) +
    H2O(c)
    average volume of E used = 19.70+19.30+19.40/3
    =19.47cm^3
    =========

    (2a)
    Inference: SO3^2- suspected
    (2b)
    (i) Inference: insoluble salt
    (ii) Inference: SO3^2- confirmed
    (iii) Inference: Cu2 suspected Cu2 confirmed
    (iv) Obseravation: -Blue gelatineous precipitate -
    Precipitate dissolves
    (2c)
    (i) CuSO3
    (ii) Copper
    (ii) trioxosulphate (iv)
    =====================

    (3ai)
    SO2(g)
    (3aii)
    A->To dry the gas
    B->To detect the type of gas that was dried
    (3aiii)
    A->P4O10
    B->KMno4/Blue litmus paper
    (3bi)
    -It is an acid that turns moist blue litmus paper
    red
    -It is colourless and poisonous gas with an
    irritating smell
    (3bII)
    The irritating smell
    (3bii)
    Downward delivery
    Chlorine gas (Cl2(g))
    (3ci)
    I-T,U,V
    II-X,Y,Z
    (3cii)
    W
    (3ciii)
    I-U
    II-V
    ++++++++++++++++++++++++++++++
    (1ai)
    CE=0.905mol/dm^3
    VD=20.0cm^3
    For the first titration
    Burette reading|1|2|3
    Final reading|39.25|29.80|19.40
    Initial reading|19.55|10.50|0.00
    Volume|19.70|19.30|19.40
    (1aii)
    Volume of E used=(2nd+3rd)titre/2
    =(19.30+19.40)/2
    =(38.70/2)cm^3
    Volume of E used=19.35cm^3
    Hence VE=19.35cm^3
    (1bi)
    From the information given,
    nD=1, nE=2,
    CD=?, CE=0.0905mol/dm^3
    VD=20.0cm^3, VE=19.35cm^3
    Using: CDVD/CEVE=nD/nE
    CD=CEVEnD/VDnE
    =(0.0905*19.35*1)/(20.0*2)
    =1.751175/40
    =0.04377
    CD=0.0438mol/dm^3
    (1bii)
    Since 0.04377mol is in 1000cm^3
    xmol will be 350cm^3
    x=0.04377*35.0/1000
    =1.5322/1000
    =0.00153mol/dm^3 at 25C
    Hence the solubility is 0.00153mol/dm^3
    (1biii)
    From the stoichiometry
    1 mole of Na2CO3 produces 1 mole of CO2
    0.00153mol of Na2CO3 produces 0.00153mole of
    CO2
    mass=molar mass*number of mole
    molar mass Na2CO3=23*2+12+16*3
    =46+12+48
    =106g/mol
    mass of CO2=12+16*2
    =12+32
    =44g/mol
    106g Na2CO3 produces 44g of CO2
    0.0153*106 will produce y
    y=0.153*106*44/106
    =0.153*44
    y=6.732g of CO2
    Hence mass of CO2 produced is 6.732g
    =======================

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