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化學混合物問題(要計數)

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發問:

有一個9g的mixture有 Al2(SO4)3.5H2O 同 Fe2(SO4)3.7H2O,但加熱後,水H2O就無晒得返7g 咁佢要搵返Al2(SO4)3.5H2O 同 Fe2(SO4)3.7H2O o既百分比. 提示係大約80%比20% 有無高手可以計到,資料得咁多 我計極都計唔到,希望各位可以幫幫我 thank you gratitude for your help 更新: 但ans係話大約80%比20% thx 幫幫我

最佳解答:

Ans: 9-7g = 2gm 2 gm is H2O No. of mole of water removed = 2/18.016 = 0.111012 let x = weight of Al2(SO4)3.5H2O then 9-x = weight of Fe2(SO4)3.7H2O no of mole of Al2(SO4)3.5H2O = x/432.22 no. of mole of Fe2(SO4)3.7H2O = (9-x)/525.972 no of mole of water formed by Al cpd is (5x)/432.22 no. of mole of water formed by Fe cpd is (63-7x)/525.972 Total no. of mole of water form is 5x/432.22 + (63-7x)/525.972 = 0.111012 -> 0.011568x + 0.119778 - 0.0133087x = 0.111012 ->0.0017407x = 0.008766 -> x = 5 Al2(SO4)3.5H2O is 5/9 = 55.56% Fe2(SO4)3.7H2O is 4/9 = 44.44%

其他解答:

寫條chemical equation出黎睇mole ratio先.. aX.5H2O (add) bY.7H2O --> aX (add) bY (add) cH2O (X=Al2(SO4)3;Y=Fe2(SO4)3) 求百分比,即係求兩者既mass,兩個unknowns,即係要solve兩條equations Molecular Weight, M of X=342.3; Y=399.9; X.5H2O=432.3; Y.7H2O=525.9 未加熱前,total mass=9g 432.3 a (add) 525.9 b = 9 --(1) 加完熱,剩番7g 342.3 a (add) 399.9 b = 7 --(2) by solving (1),(2), a=1.153*10^-2 mol b=7.632*10^-3 mol mass of X.5H2O = 432.3 a = 4.984g mass of Y.7H2O = 525.9 b = 4.014g mass ratio X.5H2O:Y.7H2O = 55% : 45%34CDDE348BEB263C