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Sebanyak 2,54 gram Cu direaksikan dengan 100 mL HNO 3 ​ 1 M menurut reaksi berikut. Cu + 4 HNO 3 ​ → Cu ( NO 3 ​ ) 2 ​ + 2 NO 2 ​ + 2 H 2 ​ O Tentukan: a. pereaksi pembatas, b. volume pereaksi tersisa jika diukur dalam keadaan STP, dan c. massa masing-masing produk yang dihasilkan.

Sebanyak 2,54 gram Cu direaksikan dengan 100 mL  menurut reaksi berikut.

 

Tentukan:

a. pereaksi pembatas,

b. volume pereaksi tersisa jika diukur dalam keadaan STP, dan

c. massa masing-masing produk yang dihasilkan.space space space

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produk yang dihasilkan sejumlah .

produk yang dihasilkan sejumlah 4 comma 69 space g space Cu open parentheses N O subscript 3 close parentheses subscript 2 semicolon space 2 comma 3 space g space N O subscript 2 semicolon space dan space 0 comma 9 space g space H subscript 2 O.space space space

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Pembahasan

a. Pereaksi pembatas Oleh karena , maka pereaksi pembatasnya adalah . b. Volume pereaksi tersisa jika diukur dalam keadaan STP Jadi, volume pereaksi tersisa adalah 0,336 L Cu dalam keadaan STP. c. Massa masing masing produk yang dihasilkan Jadi, produk yang dihasilkan sejumlah .

a. Pereaksi pembatas

n thin space Cu equals fraction numerator m space Cu over denominator italic A subscript r space Cu end fraction n thin space Cu equals fraction numerator 2 comma 54 space g over denominator 63 comma 5 space g space mol to the power of negative sign 1 end exponent end fraction n thin space Cu equals 0 comma 04 space mol  fraction numerator n space Cu over denominator koefisien space Cu end fraction equals fraction numerator 0 comma 04 space mol over denominator 1 end fraction fraction numerator n space Cu over denominator koefisien space Cu end fraction equals 0 comma 04 space mol   n thin space H N O subscript 3 double bond M cross times V n thin space H N O subscript 3 equals 1 space M cross times 0 comma 1 space L n thin space H N O subscript 3 equals 0 comma 1 space mol  fraction numerator n space H N O subscript 3 over denominator koefisien space H N O subscript 3 end fraction equals fraction numerator 0 comma 1 space mol over denominator 4 end fraction fraction numerator n space H N O subscript 3 over denominator koefisien space H N O subscript 3 end fraction equals 0 comma 025 space mol  
 

Oleh karena fraction numerator n space H N O subscript 3 over denominator koefisien space H N O subscript 3 end fraction less than fraction numerator n space Cu over denominator koefisien space Cu end fraction, maka pereaksi pembatasnya adalah H N O subscript 3.


b. Volume pereaksi tersisa jika diukur dalam keadaan STP

space space space space space space space space space space space Cu space space space space space space plus space space space 4 H N O subscript 3 space space space end subscript yields space space space Cu open parentheses N O subscript 3 close parentheses subscript 2 space space space space plus space space space space 2 N O subscript 2 space space space space plus space space space 2 H subscript 2 O M space space space space space 0 comma 04 space mol space space space space space space 0 comma 1 space mol space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space space space space space minus sign space space space space space space space space space space space space space space space space minus sign bottom enclose R minus sign begin inline style 1 fourth end style left parenthesis 0 comma 1 space mol right parenthesis space space minus sign 0 comma 1 space mol space space space space space plus begin inline style 1 fourth end style left parenthesis 0 comma 1 space mol right parenthesis space space space plus begin inline style 2 over 4 end style left parenthesis 0 comma 1 space mol right parenthesis space plus begin inline style 2 over 4 end style left parenthesis 0 comma 1 space mol right parenthesis end enclose S space space space space 0 comma 015 space mol space space space space space space space space space minus sign space space space space space space space space space space space space space space space 0 comma 025 space mol space space space space space space space space space space 0 comma 05 space mol space space space space space space space 0 comma 05 space mol  

Pereaksi space tersisa double bond Cu  V space Cu space open parentheses STP close parentheses double bond n space Cu cross times 22 comma 4 space L space mol to the power of negative sign 1 end exponent V space Cu space open parentheses STP close parentheses equals 0 comma 015 space mol cross times 22 comma 4 space L space mol to the power of negative sign 1 end exponent V space Cu space open parentheses STP close parentheses equals 0 comma 336 space L  


Jadi, volume pereaksi tersisa adalah 0,336 L Cu dalam keadaan STP.
 

c. Massa masing masing produk yang dihasilkan

1. space Cu open parentheses N O subscript 3 close parentheses subscript 2 space space space space m space Cu open parentheses N O subscript 3 close parentheses subscript 2 equals n thin space Cu open parentheses N O subscript 3 close parentheses subscript 2 cross times italic M subscript r space Cu open parentheses N O subscript 3 close parentheses subscript 2 space space space space m space Cu open parentheses N O subscript 3 close parentheses subscript 2 equals 0 comma 025 space mol cross times 187 comma 5 space g space mol to the power of negative sign 1 end exponent space space space space m space Cu open parentheses N O subscript 3 close parentheses subscript 2 equals 4 comma 69 space g  2. space N O subscript 2 space space space space m space N O subscript 2 double bond n thin space N O subscript 2 cross times italic M subscript r space N O subscript 2 space space space space m space N O subscript 2 equals 0 comma 05 space mol cross times 46 space g space mol to the power of negative sign 1 end exponent space space space space m space N O subscript 2 equals 2 comma 3 space g  3. space H subscript 2 O space space space space m space H subscript 2 O double bond n thin space H subscript 2 O cross times italic M subscript r space H subscript 2 O space space space space m space H subscript 2 O equals 0 comma 05 space mol cross times 18 space g space mol to the power of negative sign 1 end exponent space space space space m space H subscript 2 O equals 0 comma 9 space g 


Jadi, produk yang dihasilkan sejumlah 4 comma 69 space g space Cu open parentheses N O subscript 3 close parentheses subscript 2 semicolon space 2 comma 3 space g space N O subscript 2 semicolon space dan space 0 comma 9 space g space H subscript 2 O.space space space

Latihan Bab

Hitungan Stoikiometri Reaksi Sederhana

Hitungan Stoikiometri Reaksi dengan Pereaksi Pembatas

Hitungan Stoikiometri Reaksi dengan Senyawa Hidrat

Stoikiometri Senyawa

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