PTHH: \(ZnO+H_2-t^o->Zn+H_2O\)
----------0,1-----0,1---------------0,1-----0,1 ( mol )
Ta có nZnO= 8,1/81 = 0,1 mol
Theo Ptth => nH2 =0,1 mol
=> VH2 =0,1 .22,4 = 2,24 lít
Theo PTHH ta có nZn = 0,1 mol
=> mZn =0,1.65=6,5 gam
\(Zn+2HCl->ZnCl_2+H_2\)
mHCl =200.7,3%= 14,6 gam
ta có nHCl = 14,6/36,5=0,4 mol
Ta có \(\dfrac{0,1}{1}< \dfrac{0,4}{2}=>HCl,dư\)
Theo PTHH ta có nH2 = nZn =0,1 mol
=> VH2 =0,1 .22,4 =2,24 lít
Vậy...
Ta có: \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
PTHH: ZnO + H2 -to-> Zn + H2O (1)
a, Ta có: \(n_{H_2}=n_{Zn}=n_{ZnO}=0,1\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
b, \(m_{Zn}=0,1.65=6,5\left(g\right)\)
c, PTHH: Zn + 2HCl -> ZnCl2 + H2 (2)
Theo các PTHH và đề bài, ta có:
\(n_{Zn\left(2\right)}=n_{Zn\left(1\right)}=0,1\left(mol\right)\\ m_{HCl}=\dfrac{7,3.200}{100}=14,6\left(g\right)\\ =>n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Ta có: \(\dfrac{n_{Zn\left(đề\right)}}{n_{Zn\left(PTHH\right)}}=\dfrac{0,1}{1}< \dfrac{n_{HCl\left(đề\right)}}{n_{HCl\left(PTHH\right)}}=\dfrac{0,4}{2}\)
=> Zn hết, HCl dư.
=> \(n_{H_2\left(2\right)}=n_{Zn\left(2\right)}=0,1\left(mol\right)\\ =>V_{H_2\left(2,đktc\right)}=0,1.22,4=2,24\left(l\right)\)
PTHH: ZnO + H2 -to-> Zn + H2O (1)
\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
a, \(n_{H_2}=n_{Zn}=n_{ZnO}=0,1\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,m_{Zn}=0,1.65=6,5\left(g\right)\\ c,PTHH:Zn+2HCl->ZnCl_2+H_2\left(2\right)\)
Theo PTHH và đề bài, ta có: \(m_{HCl}=\dfrac{7,3.200}{100}=14,6\left(g\right)\\ =>n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\\ n_{Zn\left(2\right)}=n_{Zn\left(1\right)}=0,1\left(mol\right)\\ =>\dfrac{n_{HCl\left(đề\right)}}{n_{HCl\left(PTHH\right)}}=\dfrac{0,4}{2}>\dfrac{n_{Zn\left(đề\right)}}{n_{Zn\left(PTHH\right)}}=\dfrac{0,1}{1}\)
=> HCl dư, Zn hết, tính theo nZn
=> \(n_{H_2}=n_{Zn\left(2\right)}=0,1\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
