Ion ,mole concentration precipitate silver ions questions ?

Aqueous XO4^-3 ion forms a precipitate with aqueous Ag ion.When 41 cm^3 of solution of silver aqueous ions of concentration of 0.2 mol/dm^3

were added to the solution then 1.2 gram of precipitate is formed

a) Calculate the mole of silver ion used and 

b) calculate the amount of mole of Precipitate formed 

c)Find Molar mass of Precipitate d)Determine the RAM of X and identify X .

2 Answers

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  • 4 weeks ago
    Favorite Answer

    a)

    XO4{3-} + 3 Ag{+} → Ag3XO4(s)

    (0.041 dm^3) x (0.2 mol Ag{+}/dm^3) = 0.0082 mol Ag{+}

    b)

    (0.0082 mol Ag{+}) x (1 mol Ag3XO4(s) / 3 mol Ag{+}) = 0.0027 mol Ag3XO4(s)

    c)

    (1.2 g Ag3XO4(s)) / ((0.0082 mol Ag{+}) x (1 mol Ag3XO4(s) / 3 mol Ag{+})) =

    439 g/mol

    d)

    (439 g Ag3XO4/mol) - ((107.8682 g Ag/mol) x 3) - ((15.99943 g O/mol) x 4) =

    51.40 g X / mol

    That atomic mass is half-way between V and Cr.

    The CrO4 anion has a charge of 2-, so it does not fit this question.

    The VO4 anion has a charge of 3-, so V is the answer to this question.

  • ?
    Lv 7
    4 weeks ago

    3Ag+ + XO4^-3 = Ag3(XO4)

    moles of Ag+ ni the reaction =   0.2 mol/dm^3 * 0.041 dm^3 = 0.0082 moles 

    from the equation we see 

    moles of ppte = 0.0082 moles / 3 = 0.002733

    mass of precipitate = 1.2g so molar mass of ppte= 1.2 g / .002733 moles 

    molar mass = 439.0 g / mol of precipitate

    looking at 1 mole of Ag3(XO4) 

    atomic mass of Ag = 107.86 g/ mol 

    so 3 * 107.86 + X + 4*16 =  439.0 g / mol

    323.58 +X+ 64 = 439.0

    X = 439.0 - 64- 323.58 =  51.42 ATM

    Chromium ( 52.00) 

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