\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{HCl\left(db\right)}=0,2mol;n_{Zn\left(db\right)}=0,2mol\)
\(\)Ta có: \(\frac{n_{HCl\left(db\right)}}{2}< \frac{n_{Zn\left(db\right)}}{1}\)\(\Rightarrow\)HCl hết tính theo HCl.
Tính theo PTHH ta có: \(n_{H_2}=\frac{1}{2}n_{HCl\left(db\right)}=0,1mol\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24l\)
Vậy thể tích H2 là 2,24l.
Số mol Zn:
.........13/65=0,2(mol)
Số mol HCl:
.............7,3/36,5=0,2(mol)
Zn+ 2HCl----> ZnCl2+ H2
Có: 13/65>0,2/2=> Zn dư.
=> Tính theo HCl.
Zn+ 2HCl----> ZnCl2+ H2
0,1....0,2.........................0,1(mol)
VH2=0,1.22,4=2,24(l)
#Walker
\(n=\frac{m}{M}\)
\(\Leftrightarrow n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
\(\Leftrightarrow n_{HCl}=\frac{7,3}{36,5}=0,2\left(mol\right)\)
Ta có phương trình hóa học
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
0,1mol 2mol 0,1 mol
\(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
Số mol Zn:
.........\(\frac{13}{27}\left(mol\right)\)
Số mol HCl:
.............\(\frac{7,3}{36,5}=0,2\left(mol\right)\)
Zn+ 2HCl----> ZnCl2+ H2
Có: \(\frac{13}{27}< \frac{0,2}{2}\)=> HCl dư.
=> Tính theo Zn
Zn+ 2HCl----> ZnCl2+ H2
\(\frac{13}{27}\)..................................\(\frac{13}{27}\left(mol\right)\)
\(V_{H_2}=\frac{13}{27}.22,4=\frac{1456}{135}=10,7\left(851\right)\left(l\right)\)
#Walker
