Soal Rumus Hidrat

One response to “Soal Rumus Hidrat

  1. kinta dwi fitriani rahayu
    X.7
    1. CaCl₂ + xH₂O → CaCl₂ + xH₂O
    2gr 1,94gr 3,94gr
    (massa xH₂O = massa CaCl xH₂O-massa CaCl₂)
    3,94-2= 1,94gr
    Mol CaCl₂= massa CaCl₂/mrCaCl= 2 gr
    Mol xH₂O= massa xH₂O/x.mrH₂O= 1,94 gr/x18gr/mol= 1,94/18xmol
    Perbandingan mol CaCl₂ = mol xH₂O= 2/111: 1,94/18x
    18x.2 = 111. 1,94
    36x = 215,94
    X= 215,34/36= 6
    Rumus garam hidrat CaCl₂6H₂O
    2. X H₂O = 14,5%
    Na₂CO₃ = 85,5
    Mol xH₂O = massa x H₂O/xMrH₂O
    = 14,5 gr/18gr/mol = 0,9 mol
    Mol Na₂CO₃= massa Na₂CO₃/MrNa₂CO₃= 85,5 gr /94 gr/mol= 0,9 mol
    Na₂CO₃/xH₂O= 0,9/0,9= 1
    Rumus garam hidrat Na₂CO₃H₂O
    3. Massa H₂O = 15-8,5= 6,8 gram
    Mol FeSO₄ = 0,05
    Mol H₂O= 6,8/18= 0,36
    FeSO₄. xH₂O → FeSO₄ + xH₂O
    0,06 + 0,36
    1 : 6
    X=6
    Rumus garam hidrat FeSO₄6H₂O.

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