Chọn C

Chọn C

- Đổ dd NaOH dư vào từng lọ

+) Kim loại tan dần và có khí thoát ra: Al

PTHH: \(Al+NaOH+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\uparrow\)

+) Không hiện tượng: Mg

\(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)  (3:2:1)

\(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)  (2:3:1:6)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)  (2:2:2:1)

\(CaO+SO_2\rightarrow CaSO_3\)  (1:1:1)

Chọn C

Bài 4.3

PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)

            \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

Ta có: \(n_P=\dfrac{m}{31}\left(mol\right)\) \(\Rightarrow n_{P_2O_5}=\dfrac{m}{62}\left(mol\right)\) \(\Rightarrow n_{H_3PO_4\left(tạo.ra\right)}=\dfrac{m}{31}\left(mol\right)\) \(\Rightarrow m_{H_3PO_4\left(tạo.ra\right)}=\dfrac{98m}{31}\left(g\right)\)

Mặt khác: \(\left\{{}\begin{matrix}m_{P_2O_5}=\dfrac{142m}{62}\left(g\right)\\m_{H_3PO_4\left(ban.đầu\right)}=500\cdot20\%=100\left(g\right)\end{matrix}\right.\)

\(\Rightarrow C\%_{H_3PO_4\left(sau\right)}=\dfrac{100+\dfrac{98m}{31}}{500+\dfrac{142m}{62}}=\dfrac{40}{100}\) \(\Rightarrow m\approx44,54\left(g\right)\) 

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,15\left(mol\right)=n_{H_2}\\n_{Al_2\left(SO_4\right)_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)  (1:3:2)

\(2C_2H_2+5O_2\xrightarrow[]{t^o}4CO_2+2H_2O\)  (2:5:4:2)

\(2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)  (2:1:3)

\(4Na+O_2\rightarrow2Na_2O\)  (4:1:2)

\(2Mg+O_2\xrightarrow[]{t^o}2MgO\)  (2:1:2)

\(2Fe+3Cl_2\xrightarrow[]{t^o}2FeCl_3\)  (2:3:2)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)  (1:2:1:1)

\(2C_4H_{10}+13O_2\xrightarrow[]{t^o}8CO_2+10H_2O\)  (2:13:8:10)

\(3NaOH+AlCl_3\rightarrow3NaCl+Al\left(OH\right)_3\)  (3:1:3:1)

\(3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\)  (3:8:3:2:4)

- Dùng kim loại Nhôm để làm sạch dd AlCl₃

PTHH: \(2Al+3CuCl_2\rightarrow2AlCl_3+3Cu\)

            \(2Al+3ZnCl_2\rightarrow2AlCl_3+3Zn\)

PTHH: \(Zn+\dfrac{1}{2}O_2\xrightarrow[]{t^o}ZnO\)

Bảo toàn khối lượng: \(m_{O_2}=m_{ZnO}-m_{Zn}=1,6\left(g\right)\)