Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
PTHH: Zn + 2HCl -> ZnCl2 + H2 (1)
a) Ta có: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)
b) PTHH: 3H2 + Fe2O3 -> 2Fe + 3H2O (2)
Ta có: \(\left\{{}\begin{matrix}n_{H_2\left(2\right)}=n_{H_2\left(1\right)}=0,3\left(mol\right)\\n_{Fe_2O_3\left(2\right)}=\dfrac{19,2}{160}=0,12\left(mol\right)\end{matrix}\right.\)
Ta có: \(\dfrac{0,3}{3}< \dfrac{0,12}{1}\)
=> Fe2O3 dư, H2 hết nên tính theo H2.
=> \(n_{Fe}=\dfrac{2.0,3}{3}=0,2\left(mol\right)\\ =>m_{Fe}=0,2.56=11,2\left(g\right)\)
a) PTHH: Zn + 2HCl --> ZnCl2 + H2
Ta có: nZn = \(\dfrac{19,5}{65}\) = 0,3 mol'
Theo PT: nZn = nH2 = 0,3 mol
=> VH2 = 0,3 x 22,4 = 6,72l
b) PTHH: Fe2O3 + 3H2 --> 2Fe + 3H2O
\(n_{Fe_2O_3}\) = \(\dfrac{19,2}{160}\) = 0,12 mol
Vì \(\dfrac{0,12}{1}>\dfrac{0,3}{3}\) => Fe2O3 dư
Cứ 3 mol H2 --> 2 mol Fe
0,3 mol --> 0,2 mol'
=> mFe = 0,2 x 56 = 11,2g
PTHH : Zn + 2HCl -> ZnCl2 + H2
a , nZn = \(\dfrac{19,5}{65}=0,3\left(mol\right)\)
Theo PTHH , nZn = nH2 = 0,3 (mol)
=> VH2(đktc) = 0,3.22,4 =6,72 (l)
b , PTHH : Fe2O3 + 3H2 -> 2Fe + 3H2O
nFe2O3 = \(\dfrac{19,2}{160}=0,12\left(mol\right)\)
Vì 0,12> 0,3/3 <=>0,12>0,1 => Fe2O3 dư , H2 hết => tính theo số mol H2
Theo PTHH , nFe =\(\dfrac{2}{3}n_{H2}\) = 0,3 (mol)
=>mFe = 0,3 . 56 = 16,8 (g)
a) Có: \(n_{Zn}=\frac{19,5}{65}=0,3\left(mol\right)\)
PTHH:
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0,3-------->0,3\)
Thep phương trình trên, \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b) \(n_{Fe_2O_3}=\frac{19,2}{160}=0,12\left(mol\right)\)
PTHH:
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
\(0,3-->0,2\)
Theo phương trình,. ta thấy Fe2O3 còn dư sau phản ứng
\(\Rightarrow n_{Fe}=\frac{2}{3}n_{H_2}=\frac{2}{3}.0,3=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)
