\(n_{Zn}=\frac{m}{M}=\frac{23.5}{65}=0,36mol\)
\(m_{HCl}=\frac{200.7,3}{100}=14,6g\)
\(n_{HCl}=\frac{m}{M}=\frac{14,6}{36,5}0,4mol\)
PTHH:
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
bd 0,36 0,4 0 0(mol)
pứ 0,2 0,4 0,2 0,2(mol)
spứ 0,16 0 0,2 0,2 (mol)
a)\(V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)
b)\(m_{ZnCl_2}=n.M=0,2.136=27,2g\)
\(m_A=m_{Znpu}+m_{ddHCl}-m_{H_2}=0,2.65+200-0,2.2=212,6g\)\(C\%_{ZnCl_2}=\frac{27,2}{212,6}.100\%=12,79\%\)
a) Zn +2HCl---->ZnCl2 +H2
n\(_{Zn}=\frac{23,5}{65}=0,36\left(mol\right)\)
n\(_{HCl}=\frac{200.7,3}{100.36,5}=0,4\left(mol\right)\)
=> Zn dư( Do tỉ lệ hệ số cân bằng)
Theo pthh
n\(_{H2}=0,2\Rightarrow m_{H2}=0,4\left(g\right)\)
VH2=0,2.22,4=4,48(l)
b)mdd=200+23,5-0,4=223,1(g)
n\(_{ZnCl2}=\frac{1}{2}n_{HCl}=0,2\left(mol\right)\)
C%ZnCl2=\(\frac{0,2.136}{223,1}.100\%=12,19\%\)
BÀI 1: Cho 4,93g hỗn hợp A gồm Mg, Zn vào 250g ddHCl 7,3%
