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Kristal asam oksalat () yang massanya 5,04 gram di...

Kristal asam oksalat (H subscript 2 C subscript 2 O subscript 4 point nH subscript 2 O) yang massanya 5,04 gram dilarutkan dalam air. Larutan yang terjadi direaksikan dengan larutan Ca Cl subscript 2 sesuai reaksi:

H subscript 2 C subscript 2 O subscript 4 point nH subscript 2 O open parentheses italic s close parentheses and Ca Cl subscript 2 left parenthesis italic a italic q right parenthesis yields Ca C subscript 2 O subscript 4 left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses and H C I open parentheses italic l close parentheses 

Dari reaksi tersebut ternyata didapatkan 5,12 gram Ca C subscript 2 O subscript 4. Jika italic A subscript r Ca=40, C=12, H=1, O=16, rumus air kristalnya yang tepat adalah ...

Jawaban:

Persamaan reaksi:

H subscript 2 C subscript 2 O subscript 4 point nH subscript 2 O open parentheses italic s close parentheses and Ca Cl subscript 2 left parenthesis italic a italic q right parenthesis yields Ca C subscript 2 O subscript 4 left parenthesis italic a italic q right parenthesis plus H subscript 2 O open parentheses italic l close parentheses and 2 H C I open parentheses italic l close parentheses 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space H subscript 2 C subscript 2 O subscript 4 end cell equals cell left parenthesis 2 cross times italic A subscript r space H right parenthesis plus left parenthesis 2 cross times italic A subscript r space C right parenthesis plus left parenthesis 4 cross times italic A subscript r space O right parenthesis end cell row blank equals cell left parenthesis 2 cross times 1 right parenthesis plus left parenthesis 2 cross times 12 right parenthesis plus left parenthesis 4 cross times 16 right parenthesis end cell row blank equals cell 2 plus 24 plus 64 end cell row blank equals cell 90 space g space mol to the power of negative sign 1 end exponent end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space H subscript 2 O end cell equals cell left parenthesis 2 cross times italic A subscript r space H right parenthesis plus left parenthesis 1 cross times italic A subscript r space O right parenthesis end cell row blank equals cell left parenthesis 2 cross times 1 right parenthesis plus left parenthesis 1 cross times 16 right parenthesis end cell row blank equals cell 2 plus 16 end cell row blank equals cell 18 space g space mol to the power of negative sign 1 end exponent end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space Ca C subscript 2 O subscript 4 end cell equals cell left parenthesis 1 cross times italic A subscript r space Ca right parenthesis plus left parenthesis 2 cross times italic A subscript r space C right parenthesis plus left parenthesis 4 cross times italic A subscript r space O right parenthesis end cell row blank equals cell left parenthesis 1 cross times 40 right parenthesis plus left parenthesis 2 cross times 12 right parenthesis plus left parenthesis 4 cross times 16 right parenthesis end cell row blank equals cell 40 plus 24 plus 64 end cell row blank equals cell 128 space g space mol to the power of negative sign 1 end exponent end cell end table 

Pada reaksi diperoleh jumlah mol Error converting from MathML to accessible text. dengan H subscript 2 C subscript 2 O subscript 4 point nH subscript 2 O  sama, sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript H subscript 2 C subscript 2 O subscript 4 point nH subscript 2 O end subscript end cell equals cell n subscript Ca C subscript 2 O subscript 4 end subscript end cell row cell fraction numerator m subscript H subscript 2 C subscript 2 O subscript 4 point nH subscript 2 O end subscript over denominator italic M subscript r italic space H subscript italic 2 C subscript italic 2 O subscript italic 4 point nH subscript italic 2 O end fraction end cell equals cell fraction numerator m subscript Ca C subscript 2 O subscript 4 end subscript over denominator italic M subscript r italic space Ca C subscript italic 2 O subscript italic 4 end fraction end cell row cell fraction numerator 5 comma 04 over denominator 90 plus 18 n end fraction end cell equals cell fraction numerator 5 comma 12 over denominator 128 end fraction end cell row cell 5 comma 04 end cell equals cell 0 comma 04 cross times left parenthesis 90 plus 18 n right parenthesis end cell row cell 5 comma 04 end cell equals cell 3 comma 6 plus 0 comma 72 n end cell row cell 0 comma 72 n end cell equals cell 1 comma 44 end cell row n equals 2 end table   

Maka rumus air kristalnya adalah H subscript 2 C subscript 2 O subscript 4 point 2 H subscript 2 O 

Maka rumus air kristalnya yang tepat adalah H subscript bold 2 C subscript bold 2 O subscript bold 4 bold point bold 2 H subscript bold 2 O.

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