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標題:
Limiting Reagent and % Yield
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最佳解答:
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
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Limiting Reagent and % Yield
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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.最佳解答:
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
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