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9 October, 12:56

What volume of h2 (g) at 27oc and 680 torr is produced by the reaction of 15.0 g al metal with excess hcl (aq) to produce alcl3 and h2 (hint: you may want to write and balance an equation first) ?

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  1. 9 October, 13:20
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    Step 1:

    The balanced chemical equation for the above problem can be written as:

    2Al + 6HCl → 2AlCl₃ + 3H₂

    Step 2:

    Since 15 grams of Al metal react with the excess of HCl, therefore Al will be the limiting reactant.

    The moles of Al that reacted will be = 15/27

    =0.55 moles of Al.

    Step 3:

    Molar ratio of Al and H₂ will be 2:3.

    Step 4:

    Therefore moiles of H2 produced by 0.55 moles of Al

    0.55 moles of Al * (3 moles of H₂ : 2 moles of Al)

    =0.55*1.5

    =0.825 moles of H₂.

    Step 5:

    By using the ideal gas equation we can obtain the volume of H2 produced.

    pV=nRT (Ideal Gas Equation)

    V = (nRT) : p

    Step 6:

    n=0.825 moles

    R=0.0821 litre atm mol⁻¹ K⁻¹

    T=27 degree celsius = 27+273 K = 300 K.

    p = 680 torr * (1 atm : 760 torr)

    =0.894 atm.

    Step 7:

    V=[0.0825 moles * 0.0821 atm L mol⁻¹ K⁻¹*300 K] : 0.894 atm

    V=2.27*10⁻³ litres.
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