\(13.a)n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\\Fe+H_2SO_4\rightarrow FeSO_4+H_2\left(1\right)\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\left(2\right)\\n_{Fe}= n_{H_2}=n_{H_2SO_4\left(1\right)} =0,05mol\\m_{Fe}=0,05.56=2,8g\\ m_{Fe_2O_3}=6,8-2,8=4g \\b)n_{Fe_2O_3}=\dfrac{4}{160}=0,025mol\\ n_{H_2SO_4\left(2\right)}=0,025.3=0,075mol\\ m_{H_2SO_4}=\left(0,05+0,075\right).98=12,25g\\ m_{H_2SO_4\left(dư\right)}=12,25.10\%=1,225g\\ C_{\%H_2SO_4}=\dfrac{12,25+1,225}{100}\cdot100=13,475\%\)
14
\(a)Fe+H_2SO_4→FeSO_4+H_2(1) Fe_2O_3+3H_2SO_4→Fe_2(SO4)_3+3H_2O(2)\)
\(\left(b\right)n_{H_2}=\dfrac{2,479}{24,79}=0,1mol\\ n_{H_2}=n_{Fe}=0,1mol\\ \%m_{Fe}=\dfrac{0,1.56}{37,6}\cdot100=14,89\%\\ \%m_{Fe_2O_3}=100-14,89=85,11\%\)
\(15.\\ a.n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Zn}=a;n_{Al}=b\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}a+1,5b=0,25\\65a+27b=9,2\end{matrix}\right.\\ \Rightarrow a=b=0,1\\ \%m_{Zn}=\dfrac{0,1.65}{9,2}\cdot100=70,65\%\\ \%m_{Al}=100-70,65=29,36\%\\ b.n_{NaOH}=0,2.1=0,2mol\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl\left(dư\right)}=0,2mol\\ n_{HCl}=0,2+0,1.2+0,1.3=0,7mol\\ C_{M\left(HCl\right)}=\dfrac{0,7}{0,1}=7M\\ c.m_{dd}=100.1,2+9,2=129,2g\\ n_{ZnCl_2}=n_{Zn}=n_{Al}=n_{AlCl_3}=0,1mol\\ C_{\%ZnCl_2}=\dfrac{0,1.136}{129,2}\cdot100=10,53\%\\ C_{\%AlCl_3}=\dfrac{0,1.133,5}{129,2}\cdot100=10,33\%\)
\(C_{\%HCl}=\dfrac{\left(0,7-0,1.2-0,1.3\right).36,5}{129,2}\cdot100=5,65\%\)
