chemisty - Ans.fyi 參考答案

chemisty

用戶219436

2011-05-18 9:14 am

25cm3 of 1M solution of sodium hydroxide were place in a flask.
1.40 g of an impure specimen of ammonium chloride was added.
the flask and its contents were then carefully heated until no more ammonia gas was
evilved.theresulting solution was found to be alkaline and was diluted to exactly 250 cm3.
50cm3 of theis solution required 5.10 cm3 of 0.1M of hydrochloric acid for neutralization.
Calculate the percentage purity of the original impure ammonium chloride.

回答 (1)

知識就是力量

2011-05-19 3:03 am

✔ 最佳答案

Since the resulting solution is alkaline, the NaOH originally reacted is in excess.
NaOH + HCl ---> NaCl + H2O
No. of moles of HCl = 0.1 * 5.1/1000 = 5.1*10^-4
No. of moles of remaining NaOH in 50 cm^3 = 5.1*10^-4
No. of moles of remaining NaOH in 250 cm^3 = 5.1*10^-4 * 250/50 = 2.55*10^-3
No. of moles of NaOH reacted with NH4Cl = 1*25/1000 - 2.55*10^-3 = 0.02245
NaOH + NH4Cl ---> NaCl + NH3 + H2O
No. of moles of NH4Cl reacted = 0.02245
Mass of NH4Cl reacted = 0.02245 * (14 + 4 + 35.5) = 1.201g
Percentage purity = 1.201/1.4 * 100% = 85.79%

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