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.
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化學問題...(英文好的比較適合進來)
2009-09-21 9:02 am
回答 (1)
2009-09-21 7:16 pm
✔ 最佳答案
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
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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%
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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%
收錄日期: 2021-04-23 23:05:53
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