`#AkaShi`

PTHH:

`ZnO + 2HNO_3 -> Zn(NO_3)_2 + H_2O`

`\text{___________________________________________}`

`a)`

`->n_{ZnO}=(m_{ZnO})/(M_{ZnO})=(16,2)/81=0,2 (mol)`

`->n_{HNO_3}=(m_{dd}xxC%)/(M_{HNO_3})=(400xx15%)/63=20/21 (mol)`

Xét tỉ lệ:

`n_{ZnO}=0,2 < n_{HNO_3}=20/21:2=0,467 (mol)`

`->HNO_3` dư ; `ZnO` hết

`->n_{HNO_3}=2xxn_{ZnO}=0,2xx2=0,4 (mol)`

`->m_{HNO_3\ pư}=0,4xx63=25,2 (g)`

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`b)`

`->n_{Zn(NO_3)_2}=n_{ZnO}=0,2 (mol)`

`->m_{Zn(NO_3)_2}=n_{Zn(NO_3)_2}xxM_{Zn(NO_3)_2}=0,2xx189=37,8 (g)`

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`c)`

`->n_{HNO_3\ dư}=20/21-0,4=58/105 (mol)`

`>m_{dd\ sau\ pư}=16,2+400=416,2 (g)`

`->C%_{HNO_3}=(58/105xx63)/(416,2)xx100%=8,36%`

`->C%_{Zn(NO_3)_2}=(189xx0,2)/(416,2)xx100%=9,08%`

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`**` Đáp án:

`a) m_{HNO_3\ pư}3=25,2 (g)`

`b) m_{Zn(NO_3)_2}=37,8 (g)`

`c) C%_{Zn(NO_3)_2}=9,08%`

`C%_{HNO_3}=8,36%`

`\text{___________________________________________}`

Đáp án :

a) PTHH:Zn+2HNO3→Zn(NO3)2+H2O

mHNO3=`(400.15)/100`=60(g)

nHNO3=`60/63`≈0.9(mol)

⇒HNO3 sẽ dư

Theo pthh thì nHNO3=0.4(mol)

mHNO3=63.0,4=25.2(g)

b)Theo pthh ta có:

nZn(NO3)2=0,2(mol)

mZn(NO3)2=0,2.189=37,8(g)

c)Chất còn dư trong dung dịch sau phản ứng là HNO3 

nHNO3 dư=0.9-0,4=0,5(mol)

mHNO3 dư=0,5.63=31,5(g)

m dd sau phản ứng =16,2+400=416,2(g)

C%HNO3=`(31,5)/(416,2)`.100%≈7,5%

C%Zn(NO3)2=`(37,8)/(416,2)`.100%≈10%

Mình có làm sai thì thông cảm cho mình nha chỉ cho mình chỗ sai nữa mình ngốc lắm :<