`#AkaShi`
PTHH:
`ZnO + 2HNO_3 -> Zn(NO_3)_2 + H_2O`
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`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%`
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