Chem-percentage impurity

◆◇Dice◇◆

2012-02-24 5:22 am

有冇人可以教下我,我唔識做T_T

1. A piece of aluminium weighing 0.45g is stirred with a copper(II) solution. Find the mass of copper deposited.
eq: 2Al+3Cu2+ --- 2Al3+ +3Cu

2. 30g of an impure sample of washing soda crystals, Na2Co3．10H2O, when treated with excess of an acid give off 4.4g of carbon dioxide. What is its percentage impurity?
eq: Na2Co3．10H20+2HCL --- 2NaCl+Co2+11H2O

3. 20.85g of iron(II) sulphate-7-water crystals are strongly ignited and give 3.90g of iron(III) oxide. What is its percentage purity?
qe: 2FeSo4．7H2O---(heat) Fe2O3...

4. 2g of copper turnings are dissolved in conc. nitric acid to give a solution of copper(II) nitrate. This solution is boiled for 10 mins to destroy unreacted nitric acid, and hydrogen sulphide is passed into the cooled liquid. What is the max. mass of copper (II) sulphide that can be formed? What is the mass if 30% of the copper content have been lost in the intermediate steps?

如果可以最好可以解釋埋...T_T

回答 (1)

andrew

2012-02-24 6:23 am

✔ 最佳答案

1.
2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu

No. of moles of Al used = 0.45/27 = 0.01667 mol
No. of moles of Cu deposited = 0.01667 x (3/2) = 0.02501 mol

Mass of Cu deposited = 0.02501 x 63.5 = 1.59 g

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2.
Na2CO3•10H2O + 2HCl → 2NaCl + CO2 +11H2O

No. of moles of CO2 formed = 4.4/(12 + 16x2) = 0.1 mol
No. of moles of Na2CO3•10H2O = 0.1 mol
Mass of Na2CO3•10H2O = 0.1 x(23x2 + 12 + 16x13 + 1x20) = 28.6

% impurity = (28.6/30) x 100% = 95.3%

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3.
2FeSO4•7H2O → Fe2O3+ ......

No. of moles of Fe2O3 = 3.90/(55.8x2 + 16x3) = 0.02444mol
No. of moles of FeSO4•7H2O = 0.02444 x2 = 0.04888 mol
Mass of FeSO4•7H2O = 0.04888 x(55.8 + 32.1 + 16x11 + 1x14) = 13.58 g

% purity = (13.58/20.85) x 100% = 65.1%

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4.
Cu + ... → → CuS + ....

No. of moles of Cu = 2/63.5 mol
No. of moles of CuS = 2/63.5 mol

Max. mass of CuS = (2/63.5) x (63.5 + 32.1) = 3.01 g

If 30% of the copper content have been lost in the intermediate step,
mass of CuS = (2/63.5) x (63.5 + 32.1) x (1 - 30%) = 2.11 g

參考: andrew

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