F.4 Chemistry(Moles)

👨🏻‍🏫 學術台化學

知識就是力量

2009-11-30 2:34 am

About percentage Yield=actual yield / theoretical yield x 100%

Question:
100cm^3 of barium chloride solution of concentration 0.0500 mol dm^-3 were treated with an excess of sulphate ions in solution. The precipitate of barium sulphate formed was dried and weighed. A mass of 1.1558g was recorded. What percentage yield does this represent?

Ans:99.2%

更新1:

Show necessary steps

回答 (1)

老爺子

2009-11-30 9:45 am

✔ 最佳答案

Ba^2+(aq) + SO4^2-(aq) → BaSO4(s)
Mole ratio Ba^2+ : BaSO4 = 1 : 1

No. of moles of BaCl2 used = 0.0500 x (100/1000) = 0.005 mol
No. of moles of Ba^2+ used = 0.005 mol
Maximum no. of moles of BaSO4 formed = 0.005 mol
Molar mass of BaSO4 = 137.3 + 32.1 + 16x4 = 233.4 g/mol
Maximum mass of BaSO4 formed = 0.005 x 233.4 = 1.167 g
Percentage yield = (1.1558/1.167) x 100% = 99.0%

(The answer is not 99.2%)
(If 1.1558 g is replaced by 1.159 g, the answer will be 99.2%)

👍 0 👎 0

收錄日期: 2021-04-23 18:21:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091129000051KK01451

檢視 Wayback Machine 備份