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Stoikiometri Kimia : Contoh Soal No 11-20

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250x250
11. Sebanyak 11,6 gram senyawa hidrat Na2SO4.xH2O dipanaskan sehingga menghasilkan 7,1 gram Na2SO 4 anhidrat. Tentukanlah rumus kimia senyawa hidrat tersebut ! (Diketahui Ar Na = 23; S = 32 ; O = 16 ; H = 1)

Jawaban : Na2SO4.5H2O
 
Pembahasan :
 
Massa Na2SO4.xH2O = 11,6 gram
Massa Na2SO4 = 7,1 gram
Massa H2O = 11,6 gram – 7,1 gram = 4,5 gram

Mol Na2SO4 = Massa / Mr = 7,1 / 142 = 0,05 mol
Mol H2O = Massa / Mr = 4,5 / 18 = 0,25 mol

Na2SO4.xH2O → Na2SO4 + x.H2O

Rumus senyawa hidrat Na2SO4.5H2O

12. Sebanyak 97,6 gram senyawa hidrat BaCl2.nH2O dipanaskan sehingga menghasilkan 83,2 gram BaCl2 anhidrat . Tentukanlah rumus kimia senyawa hidrat tersebut ! (Diketahui Ar Ba = 137 ; Cl = 35,5 ; O = 16 ; H = 1)

Jawaban : BaCl2.2H2O
 
Pembahasan :
 
Massa BaCl2.nH2O = 97,6 gram
Massa BaCl2 = 83,2 gram
Massa H2O = 97,6 gram -83,2 gram = 14,4 gram

Mol BaCl2 = Massa / Mr = 83,2 / 208 = 0,4 mol
Mol H2O = Massa / Mr = 14,4 / 18 = 0,8 mol

BaCl2.xH2O → BaCl2 + x.H2O

Rumus senyawa hidrat BaCl2.2H2O

13. Sebanyak 249,5 gram CuSO4.xH2O dipanaskan sehingga menghasilkan 159,5 gram CuSO4 anhidrat. Tentukanlah nilai x pada senyawa hidrat tersebut ! (Diketahui Ar Cu = 63,5 ; S = 32 ; O = 16 ; H = 1)

Jawaban : 5
 
Pembahasan :
 
Massa CuSO4.xH2O = 249,5 gram
Massa CuSO4 = 159,5 gram
Massa H2O = 249,5 gram – 159,5gram = 90 gram

Mol CuSO4 = Massa / Mr = 159,5 / 159,5 = 1 mol
Mol H2O = Massa / Mr = 90 / 18 = 5 mol

CuSO4.xH2O → CuSO4 + x.H2O



14. Sebanyak 24,6 gram MgSO4.xH2O dipanaskan sehingga menghasilkan 12 gram MgSO4 anhidrat. Tentukanlah nilai x pada senyawa hidrat tersebut ! (Diketahui Ar Mg= 24; S = 32 ; O = 16 ; H = 1)


Jawaban : 7
 
Pembahasan :
 
Massa MgSO4.xH2O = 24,6 gram
Massa MgSO4 = 12 gram
Massa H2O = 24,6 -12 = 12,6 gram

Mol MgSO4 = Massa / Mr = 12 / 120 = 0,1 mol
Mol H2O = Massa / Mr = 12,6/ 18 = 0,7 mol

MgSO4.xH2O → MgSO4 + x.H2O

15. Sebanyak 39,807 gram senyawa hidrat NiCl2.xH2O dipanaskan sehingga menghasilkan 38,907 gram NiCl2 anhidrat. Tentukanlah rumus kimia senyawa hidrat tersebut ! (Diketahui Ar Ni = 58,69; Cl = 35,5 ; O = 16 ; H = 1)

Jawaban : NiCl2.6H2O
 
Pembahasan :
 
Massa NiCl2.xH2O = 39,807 gram
Massa NiCl2 = 38,907 gram
Massa H2O = 39,807 gram – 38,907 gram = 0,9 gram

Mol NiCl2 = Massa / Mr = 38,907 / 129,69 = 0,3 mol
Mol H2O = Massa / Mr = 0,9/ 18 = 0,05 mol

NiCl2.xH2O → NiCl2 + x.H2O

Rumus senyawa hidrat NiCl2.6H2O

16. Suatu senyawa memiliki rumus kimia X(OH)2 . Jika massa 1 atom C-12 adalah 1,99 x 10 -23 gram dan massa 1 atom X adalah 2,275 x 10-22 gram , maka berapa persen unsur oksigen dalam senyawa tersebut ? (Diketahui Ar O=16 ; H=1)

Jawaban : 18,6931 %
 
Pembahasan :

Ar X =  Massa 1 atom X / 1/12 x massa atom C-12
= 2,275 x 10-23 / 1/12 x 1,99 x 10-23
= 137,185929 
 
Mr X(OH)2 = (137,1859296) + (2x16) + (2x1)
= 137,1859296 + 32 + 2
= 171,1859296

Persen O dalam X(OH)2 = 2 x 16 / 171,1859296 x 100 % = 18,6931 %

17. Suatu senyawa memiliki rumus kimia XF . Jika massa 1 atom C-12 adalah 1,99 x 10 -23 gram dan massa 1 atom X adalah 3,815 x 10-23 gram , maka berapa persen unsur X dalam senyawa tersebut ? (Diketahui Ar F = 19)

Jawaban : 54,7673 %
 
Pembahasan :
 
Ar X =  Massa 1 atom X / 1/12 x massa atom C-12
= 3,815 x 10-23 / 1/12 x 1,99 x 10-23
= 23,00502513

Mr XF = (23,00502513) + (19)
= 42,00502513

Persen X dalam XF = 23,00502513 / 42,00502513 x 100 % = 54,7673 %


18. Suatu senyawa memiliki rumus kimia X(NO3)2 . Jika massa 1 atom C-12 adalah 1,99 x 10 -23 gram dan massa 1 atom X adalah 6,64 x 10-23 gram, maka berapa persen unsur X dalam senyawa tersebut ? (Diketahui Ar N = 19 ; O=16)


Jawaban : 24,4088%
 
Pembahasan :


Ar X =  Massa 1 atom X / 1/12 x massa atom C-12
= 6,64 x 10-23 / 1/12 x 1,99 x 10-23
= 40,04020101

Mr X(NO3)2 = (40,04020101)+ (2 x 14) + (6 x 16)
= 164,040201

Persen X dalam X(NO3)2  = 40,04020101 / 164,040201 x 100 % = 24,4088 %

19. Suatu senyawa memiliki rumus kimia HgX . Jika massa 1 atom C-12 adalah 1,99 x 10 -23 gram dan massa 1 atom unsur X adalah 2,66 x 10-23 gram, maka berapa persen unsur Hg dalam senyawa tersebut ? (Diketahui Ar Hg= 200,6)

Jawaban : 92,5959%
 
Pembahasan :

Ar X =  Massa 1 atom X / 1/12 x massa atom C-12
= 2,66 x 10-23 / 1/12 x 1,99 x 10-23
= 16,04020101

Mr HgX = (200,6) + (16,04020101)
= 216,640201

Persen Hg dalam HgX  = 200,6 / 216,640201 x 100 % = 92,5959 %

20. Suatu senyawa memiliki rumus kimia X2Cr2O 7. Jika massa 1 atom C-12 adalah 1,99x10-23 gram dan massa 1 atom X adalah 6,4675 x 10-23 gram, maka berapa persen unsur Cr dalam senyawa tersebut ? (Diketahui Ar Cr= 52; O=16)

Jawaban : 35,3741 %
 
Pembahasan :

Ar X =  Massa 1 atom X / 1/12 x massa atom C-12
= 6,4675 x 10-23 / 1/12 x 1,99 x 10-23
= 39

Mr X2Cr2O7 = (2 x 39) + (2 x 52) + (7 x 16)
= 294

Persen Hg dalam HgX  = 200,6 / 216,640201 x 100 % = 92,5959 %

Soal Selanjutnya > Soal No 21-30
300x600
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