3 Mole Problems

[匿名]

2008-06-16 2:27 am

1) A sample of a certain concentrated acid has a density of 1.96gcm^-3 and contains 95% of the acid by mass. What is the concentration of the acid in the sample?
2) 1.05g of a mixture of anhydrous sodium carbonate and sodium chloride was dissolved in 50cm^3 of deionized water. The resulting solution required 28.5cm^3 of 0.15M sulphuric acid for complete reaction. What is the percentage purity of the anhydrous sodium carbonate sample ?
3) 3.65g of HCl(g) is dissolved in 100cm^3 of distilled water. What volume of 0.2M NaOH can neutralize the resulting solution ?

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回答 (1)

Uncle Michael

2008-06-16 6:05 am

✔ 最佳答案

1)
Consider a 1 dm3 (1000 cm3) sample of the concentrated acid.
Mass of the sample = volume x density = 1000 x 1.96 = 1960 g
Mass of the acid in the sample = 1960 x 0.95 = 1862 g
Concentration of the acid = mass/volume = 1862/1 = 1862 g dm-3
(or 1.862 g cm-3)

(If the molar mass of the acid is known,)
(Molarity of the acid = 1862/(molar mass) M)

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2)
Only anhydrous Na2CO3 reacts with H2SO4.
Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2
Mole ratio Na2CO3 : H2SO4 = 1 : 1

No. of moles of H2SO4 used = MV = 0.15 x (28.5/1000) = 0.004275 mol
No. of moles of Na2CO3 used = 0.004275 mol
Molar mass of Na2CO3 = 23x2 + 12 + 16x3 = 106 g mol-1
Mass of Na2CO3 = mol x (molar mass) = 0.004275 x 106 = 0.4532 g
Percentage by mass of Na2CO3 = (0.4532/1.05) x 100% = 43.16%

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3)
HCl + NaOH → NaCl + H2O
Mole ratio HCl : NaOH = 1 : 1

Molar mass of HCl = 1 + 35.5 = 36.5 g mol-1
No. of moles of HCl used = mass/(molar mass) = 3.65/36.5 = 0.1 mol
No. of moles of NaOH used = 0.1 mol
Volume of NaOH = mol/M = 0.1/0.2 = 0.5 dm3 = 500 cm3

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