a.PTHH:
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2 (1)
b, Ta có:
nZn = \(\dfrac{13}{65}\) = 0,2 (mol)
Theo (1): nHCl = 2nZn = 2.0,2 = 0,4 mol
\(\Rightarrow\) mHCl = 0,4. 36,5 = 14,6 (g)
c, Theo (1): nZnCl2 = nZn = 0,2 mol
\(\Rightarrow\) mZnCl2 = 0,2 . 136 = 27,2 (g)
d, Theo (1): nH2 = nZn = 0,2 mol
\(\Rightarrow\) VH2 = 0,2 . 22.4 = 4,48 (l)
a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) \(n_{Zn}=\dfrac{m}{M}=\dfrac{13}{65}=0.2\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
theo pt: 1mol 2mol 1mol 1mol theo đb:0,2mol 0,4mol 0,2mol 0,2mol b) \(m_{HCl}=n.M\)\(=0,4.36,5=14,6\left(g\right)\)
c) \(m_{ZnCl_2}=n.M\)
\(=0,2.136=27,2\left(g\right)\)
d) \(V_{H_2}=n.22,4\)
\(=0,2.22,4=4,48\left(l\right)\)
