a) \(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
\(\left(mol\right)\)___\(0,1\)___\(0,2\)______\(0,1\)_____
b) \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(C\%_{HCl}=\frac{7,3}{200}.100\%=3,65\%\)
c)
\(m_{ddZnCl_2}=m_{Zn}+m_{ddHCl}-m_{H_2}=6,5+200-0,1.2=206,3\left(g\right)\)
\(C\%_{ZnCl_2}=\frac{0,1.136}{206,3}.100\%=6,59\%\)
d) \(n_{Fe_2O_3}=\frac{16}{160}=0,1\left(mol\right)\)
\(PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Lập tỉ lệ: \(\frac{0,1}{1}>\frac{0,1}{3}\)
=> Lấy số mol của H2
\(n_{Fe}=\frac{0,1.3}{2}=\frac{1}{15}\left(mol\right)\)
\(m_{Fe}=\frac{1}{15}.56=3,73\left(g\right)\)