\(n_{Fe}=\dfrac{84}{56}=1,5\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=1,5\left(mol\right)\\ V_{H_2}=1,5.22,4=33,6\left(l\right)\\ C\%_{ddFeCl_2}=\dfrac{127.1,5}{84+300-1,5.2}.100\%=\dfrac{190,5}{381}.100\%=50\%\)

làm tự luận thôi ạ!

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right);n_{H_2SO_4}=0,05.2=0,1\left(mol\right)\\ Vì:\dfrac{0,5}{1}>\dfrac{0,1}{1}\Rightarrow Fe.dư\\ n_{Fe\left(p.ứ\right)}=n_{FeSO_4}=n_{H_2}=n_{H_2SO_4}=0,1\left(mol\right)\\ m_{Fe\left(p.ứ\right)}=0,1.56=5,6\left(g\right);m_{H_2}=2.0,1=0,2\left(g\right)\\ m_{FeSO_4}=152.0,1=15,2\left(g\right);m_{H_2SO_4}=98.0,1=9,8\left(g\right)\)

\(Vì:1< \dfrac{a}{b}=\dfrac{0,2}{0,15}=\dfrac{4}{3}< 2\)

Nên SP là hỗn hợp 2 muối axit NaH₂PO₄ và Na₂HPO₄

\(BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{ZnCl_2}=\dfrac{20,6}{136}\approx0,15147\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,2}{1}>\dfrac{0,15147}{1}\\ Có:m_{ZnCl_2\left(LT\right)}=0,2.136=27,2\left(g\right)\\ H=\dfrac{20,6}{27,2}.100\%\approx75,735\%\)

\(n_{NaOH}=\dfrac{24}{40}=0,6\left(mol\right);n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ Vì:\dfrac{n_{NaOH}}{n_{CO_2}}=2\)

Nên: SP thu được duy nhất một muối trung hoà Na₂CO₃ , không chất nào dư

\(PTHH:2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\\ n_{Na_2CO_3}=n_{CO_2}=0,3\left(mol\right)\\ m_{Na_2CO_3}=106.0,3=31,8\left(g\right)\)

\(CaC_2+2H_2O\rightarrow Ca\left(OH\right)_2+C_2H_2\\ n_{CaC_2}=n_{C_2H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ m_{CaC_2}=64.0,05=3,2\left(g\right)\\ \%m_{\dfrac{CaC_2}{đất.đèn}}=\dfrac{3,2}{4}.100\%=80\%\)

\(n_{C_2H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:CaC_2+2H_2O\rightarrow Ca\left(OH\right)_2+C_2H_2\\ n_{CaC_2}=n_{C_2H_2}=0,2.64=12,8\left(g\right)\\\%m_{\dfrac{CaC_2}{đất.đèn}}=\dfrac{12,8}{16}.100\%=80\%\)