a) nZn = 0.65/65=0.01mol
nHCl = 7.3/36.5= 0.2 mol
Zn + 2HCl -> ZnCl2 + H2
(mol) 1 2
(mol) 0.01 0.2
Lập tỉ lệ: \(\dfrac{0.01}{1}< \dfrac{0.2}{2}\). Vậy HCl dư
Zn + 2HCl -> ZnCl2 + H2
(mol) 0.01 0.02 0.01
VH2 = 0.01*22.4=0.224 (l)
b) nHCl dư = 0.2-0.02 = 0.18mol
Zn + 2HCl -> ZnCl2 + H2
(mol) 0.9 0.18
Khối lượng kẽm cần bổ sung để tác dụng hết với HCl dư là:
mZn = 0.9*65=58.6(g)
\(n_{Zn\left(1\right)}=\dfrac{m}{M}=\dfrac{0,65}{65}=0,01\left(mol\right)\\ n_{HCl}=\dfrac{m}{M}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
\(\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\left(1\right)\\ Theo\text{ }pthh:1mol\text{ }\text{ }\text{ }\text{ }\text{ }2mol\\ Theo\text{ }đb:0,01mol\text{ }\text{ }\text{ }\text{ }0,2mol\\ Pứ:\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{0,01mol}{ }\text{ }\text{ }\text{ }\text{ }\text{ }\dfrac{0,02mol}{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0,01mol\\ Sau\text{ }pứ:\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }0,18mol\)
\(\Rightarrow V_{H_2}=n\cdot22,4=0,01\cdot22,4=0,224\left(l\right)\)
\(\text{b) }pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\left(2\right)\)
Theo \(pthh\left(2\right):n_{Zn\left(2\right)}=\dfrac{1}{2}n_{HCl\left(dư\right)}=\dfrac{1}{2}\cdot0,18=0,09\left(mol\right)\)
\(m_{Zn\left(2\right)}=n\cdot M=0,09\cdot65=5,85\left(g\right)\)