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17 March, 13:35

A mixture of aluminum and iron weighing 9.62 g reacts with hydrogen chloride in aqueous solution according to the parallel reactions 2 Al + 6 HCl - > 2 AlCl3 + 3 H2 Fe + 2 HCl - > FeCl2 + H2 A 0.738 g quantity of hydrogen is evolved when the metals react completely. Calculate the mass of iron in the original mixture. Suppose that in the previous reaction I also get 15.95 g AlCl3. What is the percent yield of AlCl3?

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  1. 17 March, 14:01
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    for a)

    mAl = 3.87 gr

    mFe = 5.75 gr

    for b)

    Y = 13.1%

    Explanation:

    since aluminium reacts according to

    2 Al + 6 HCl - > 2 AlCl₃ + 3 H₂

    then 2 moles of Al generates 3 moles of H2

    then the mass of hydrogen gas obtained per mass of aluminium that reacts is

    R₁ = 3 n H₂ / 2 n Al = [3 mol * 2 gr/mol] / [2 mol * 27 gr/mol ] = 1/9

    then for iron

    Fe + 2 HCl - > FeCl₂ + H₂

    the mass of hydrogen gas obtained per mass of iron # that reacts is

    R₂ = 3 n H₂ / 2 n Fe = [3 mol * 2 gr/mol] / [2 mol * 56 gr/mol ] = 3/56

    then the mass of hydrogen obtained is

    mAl*R₁ + mFe*R₂ = mH

    mAl+mFe = mM

    where m represents mass, M the total mass of the metals, H the mass of hydrogen

    thus

    mAl = mM - mFe

    and

    mAl*R₁ + (mM-mAl) * R₂ = mH

    mAl * (R₁-R₂) + mM * R₂ = mH

    mAl = (mH - mM * R₂) / (R₁-R₂)

    replacing values

    mAl = (mH - mM * R₂) / (R₁-R₂) = (0.738 g - 9.62 g*3/56) / (1/9-3/56) = 3.87 gr

    and

    mFe = mM - mAl = 9.62 g - 3.87 gr = 5.75 gr

    for b)

    since

    m AlCl3 = m AlCl3 * (m Fe / mAlCl3) = 33.87 g * (97 g / 27 g) = 121.68 g

    the yield of AlCl3 is

    Y = 15.95 g / 121.68 g = 0.131 = 13.1%
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