Anonymous
Anonymous asked in 科學化學 · 1 decade ago

化學問題...(英文好的比較適合進來)

1. An 18.6-g sample of K2CO3 was treated in such a way that all of its carbon was captured in the compund K2Zn3[Fe(CN)6]2. Compute the mass (in grams) of this product.

2. A mixture consisting of only NaCl and KCl weighs 1.0000g. When the mixture is dissolved in water and an excess of silver nitrate is added, all the chloride ions associated with the original mixture are precipitated as insoluble AgCl. The mass of AgCl is 2.1476g. Calculate the mass percentages of NaCl and KCl in the original mixture.

3. The iron Oxide Fe2O3 reacts with CO:

Fe2O3 + 3CO --> 2Fe + 3CO2

The reaction of 433.2g of Fe2O3 with excess CO yields 254.3g of iron. Calculate the theoretical yeild of iron (assuming complete reaction) and its percentage yield.

麻煩解題, 順便詳解, 謝謝!

:)

1 Answer

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  • 1 decade ago
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    1.

    Molar mass of K2CO3

    = 39.01x2 + 12.01 + 16x3

    = 138.03 g/mol

    Molar mass of K2Zn3[Fe(CN)6]2

    = 39.01x2 + 65.38x3 + 2x[55.85 + 6x(12.01 + 14.01)]

    = 698.1 g/mol

    Each formula unit of K2CO3 contains 1 C atom, and each formula unit of K2Zn3[Fe(CN)6]2 contains 12 C atoms.

    Hence, mole ratio K2CO3 : K2Zn3[Fe(CN)6]2 = 12 : 1

    No. of moles of K2CO3 = 18.6/138.03 mol

    No. of moles of K2Zn3[Fe(CN)6]2 formed = (18.6/138.03) x (1/12) mol

    Mass of K2Zn3[Fe(CN)6]2 formed

    = (18.6/138.03) x (1/12) x (698.1)

    = 7.84 g

    =====

    2.

    Let n% be the mass percentage of NaCl in the original mixture.

    Hence, mass percentage of KCl in the original mixture = (100 - n)%

    Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol

    Molar mass of KCl = 39.01 + 35.45 = 74.46 g/mol

    Molar mass of AgCl = 107.9 + 35.45 = 143.35 g/mol

    NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)

    KCl(aq) + AgNO3(aq) → AgCl(s) + KNO3(aq)

    No. of moles of NaCl used = (1 x n%)/58.44 = n/5844

    No. of moles of KCl used = [1 x (100 - n)%]/74.46 = (100 - n)/7446

    No. of moles of AgCl formed = 1/143.35

    n/5844 + (100 - n)/7446 = 2.1476/143.35

    n/5844 + 1/74.46 - n/7446 = 2.1476/143.35

    n(1/5844 - 1/7446) = 2.1476/143.35 - 1/74.46

    n = (2.1476/143.35 - 1/74.46) / (1/5844 - 1/7446)

    n = 42.14

    Mass percentage of NaCl in the original mixture = 42.14%

    Mass percentage of KCl in the original mixture = 1 - 42.14% = 57.86%

    =====

    3.

    Molar mass of Fe = 55.85 g/mol

    Molar mass of Fe2O3 = 55.85x2 + 16x3 = 159.7 g/mol

    Fe2O3 + 3CO → 2Fe + 3CO2

    Mole ratio Fe2O3 : Fe = 1 : 2

    No. of moles of Fe2O3 used = 433.2/159.7 = 2.713 mol

    No. of moles of Fe formed theoretically = 2.713 x 2 = 5.426 mol

    Theoretical yield of Fe = 5.426 x 55.85 = 303.0 g

    Percentage yield of Fe = (254.3/303.0) x 100% = 83.93%

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