Bài 5:
\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\\ 2KMnO_4+16HCl_{đặc,nóng}\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\\ n_{Cl_2\left(LT\right)}=\dfrac{5}{2}.n_{KMnO_4}=2,5.0,2=0,5\left(mol\right)\\ n_{Cl_2\left(TT\right)}=0,5.90\%=0,45\left(mol\right)\\ \Rightarrow V_{Cl_2\left(đktc\right)}=0,45.22,4=10,08\left(l\right)\)
Bài 4:
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\CuO+2HCl\rightarrow CuCl_2+H_2\\ n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ a,\%m_{Fe}=\dfrac{56.0,1}{13,6}.100\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow C_{MddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
Bài 3: Em tham khảo đây nha!
https://hoc24.vn/cau-hoi/cho-132g-hon-hop-mg-va-fe-tac-dung-voi-200ml-dd-hclsau-phan-ung-thu-duoc-784-lit-khi-h2-dktc-atinh-phan-tram-khoi-luong-moi-kim-loai-trong-hon.4697651669843
Bài 2:
\(2Na+X_2\rightarrow2NaX\\ n_{X_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ \Rightarrow n_{NaX}=0,2.2=0,4\left(mol\right)\\ \Rightarrow M_{NaX}=\dfrac{41,2}{0,4}=103\left(\dfrac{g}{mol}\right)=23+M_X\\ \Leftrightarrow M_X=80\left(\dfrac{g}{mol}\right)\\ \Rightarrow X:Br;X_2:Br_2\)
Bài 1:
\(Mg+X_2\rightarrow MgX_2\\ Áp.dụng.ĐLBTKL:m_{Mg}+m_{X_2}=m_{MgX_2}\\ \Leftrightarrow3,6+m_{X_2}=14,25\\ \Leftrightarrow m_{X_2}=10,65\left(g\right)\\ n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\Rightarrow n_{X_2}=0,15\left(mol\right)\\ \Rightarrow M_{X_2}=\dfrac{10,65}{0,15}=71\left(\dfrac{g}{mol}\right)=2.M_X\\ \Leftrightarrow M_X=35,5\left(\dfrac{g}{mol}\right)\\ \Rightarrow X:Clo\left(Cl=35,5\right)\)
