Kedalam 150 mL HCI 0,5 M dimasukkan 1,62 gram logam Aluminium, Al =27, H =1, CI =36. Tentukan : a. Persamaan reaksi. b. Pereaksi pembatas c. Berat zat yang sisa d. Volume gas Hidrogen (STP) e.

Kedalam 150 mL HCI 0,5 M dimasukkan 1,62 gram logam Aluminium, $begin mathsize 14px style A subscript italic r end style$ Al =27, H =1, CI =36. Tentukan :

a. Persamaan reaksi.

b. Pereaksi pembatas

c. Berat zat yang sisa

d. Volume gas Hidrogen (STP)

e. Volume gas hidrogen bila diukur pada suhu 27 $begin mathsize 14px style degree C end style$ dan tekanan 2 atm

Jawaban:

$begin mathsize 14px style a point space 2 Al and 6 H Cl yields Al Cl subscript 3 and 3 H subscript 2 b point space mol space Al equals massa over Ar equals fraction numerator 1 comma 62 over denominator 27 end fraction equals 0 comma 06 space mol space mol space H Cl double bond M cross times V equals 0 comma 5 cross times 0 comma 15 equals 0 comma 075 space mol Pereaksi space pembatas space colon space mol over koefisien lebih space kecil space yaitu space H Cl c point space mol space Al space bereaksi equals fraction numerator Koef space Al over denominator Koef space H Cl end fraction cross times mol space H Cl space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 2 over 6 cross times 0 comma 075 equals 0 comma 025 space mol space space space space space space space space space space space mol space Al space sisa double bond mol space awal space Al bond mol space Al space bereaksi space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 0 comma 06 minus sign 0 comma 025 equals 0 comma 035 space mol space space space space space space space space space space space space space massa space Al double bond mol cross times Ar space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 0 comma 035 cross times 27 equals 0 comma 945 space gram space d point space mol space H subscript 2 equals fraction numerator Koef space H subscript 2 over denominator Koef space H Cl end fraction cross times mol space H Cl space space space space space space space space space space space space space space space equals 3 over 6 cross times 0 comma 075 space space space space space space space space space space space space space space space equals 0 comma 0375 space mol space space space space space space space V space H subscript 2 double bond mol cross times 22 comma 4 L space space space space space space space space space space space space space space space equals 0 comma 0375 cross times 22 comma 4 space space space space space space space space space space space space space space space equals 0 comma 84 space L e point space space P V double bond nRT space space space space space space space space V equals nRT over P space space space space space space space space space space equals fraction numerator 0 comma 0375 space mol cross times 0 comma 08 space L space atm forward slash mol space K cross times 300 space K over denominator 2 space atm end fraction space space space space space space space space space space equals 0 comma 45 space L end style$

Jadi, jawaban yang benar adalah sesuai perhitungan di atas.

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