Limiting Reagent and % Yield
2012-02-19 12:50 am
Under appropriate reaction conditions Al and S produce Al2S3 according to the equation2Al + 3S → Al2S3In a certain experiment with 22.1 g of Al and excess of S, a yield of 63.4% was obtained. What was the actual yield (i.e. mass) of Al2S3 obtained.
回答 (1)
2012-02-19 7:12 am
✔ 最佳答案
Method1 :2Al + 3S → Al2S3
No. of moles of Al used = 22.1/27 = 0.8185 mol
Actual no. of moles of Al2S3 obtained = 0.8185 x(1/2) x63.4%mol
Actual yield of Al2S3 = 0.8185 x(1/2) x63.4% x(27x2 +32.1x3) = 39.0 g
Method2:
Mass fraction of Al in Al2S3 = (27x2)/(27x2 + 32.1x3)=54/150.3
Actual yield of Al2S3 = 22.1 ÷(54/150.3) x 63.4% =39.0 g
參考: sioieng
收錄日期: 2021-04-22 23:40:43
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