\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ \Rightarrow n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,1(mol)\\ c,m_{Zn}=0,1.65=6,5(g)\\ m_{ZnCl_2}=0,1.136=13,6(g)\)

Gọi hóa trị của M là \(a\)

\({M_1}^a{Cl_2}^1\Rightarrow a=2.1=2 \Rightarrow M(II)\\ CTTQ:{M_x}^{II}{O_y}^{II}\\ \Rightarrow II.x=II.y \Rightarrow \dfrac{x}{y}=1 \Rightarrow x=y=1\\ \Rightarrow MO\)

\(4P+5O_2\rightarrow P_2O_5\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\\ n_{P_2O_5}=\dfrac{1}{2}n_P=0,2\left(mol\right)\\ n_{H_3PO_4}=2n_{P_2O_5}=0,4\left(mol\right)\\ m_{ddH_3PO_4}=200.1,03=206\left(g\right)\\ \Rightarrow C\%_{H_3PO_4\left(saupu\right)}=\dfrac{0,4.98+206.5\%}{206+0,2.142}.100=21,12\%\)

\(a.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.CuO+2HCl\rightarrow CuCl_2+H_2\\ c.Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ d.MgO+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{H_2}=\dfrac{3,36}{22,4}0,15(mol)\\ a,PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ b,n_{Fe}=n_{H_2}=0,15(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,15.56}{14,8}.100\%=56,76\%\\ \Rightarrow \%_{Cu}=100\%-56,76\%=43,24\%\\ c,n_{H_2SO_4}=0,15(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,15.98}{20\%}=73,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{73,5}{1,4}=52,5(l)\)

Gọi nguyên tử khối trung bình của 2 kim loại cần tìm là R

\(R+2HCl\rightarrow RCl_2+H_2\\ n_R=n_{H_2}=0,075\left(mol\right)\\ \Rightarrow M_R=\dfrac{2,2}{0,075}=29,33\\ \Rightarrow2kimloạicầntìmlà:Mg,Ca\)

nhấn vào phần đọc tiếp

Câu đầu.

\(a,\) Gọi CT đơn giản nhất là \(C_xH_yO_z\)

\(\%m_{O}=100\%-42,11\%-6,43\%=51,46\%\\ x:y:z=\dfrac{\%_C}{12}:\dfrac{\%_H}{1}:\dfrac{\%_O}{12}=3,51:6,43:3,22\\ \Rightarrow x:y:z=12:22:11\\ \Rightarrow \text{CTĐGN của A là }(C_{12}H_{22}O_{11})_n\\ M_A=n.(12.12+22.1+11.16)=342\\ \Rightarrow n=1\\ \Rightarrow CTHH_A:C_{12}H_{22}O_{11}\)

\(b,d_{B/kk}=2,206\\ \Rightarrow M_B=2,206.29\approx 64(g/mol)\\ n_S=\dfrac{64.50\%}{32}=1(mol)\\ n_O=\dfrac{64.50\%}{16}=2(mol)\\ \Rightarrow CTHH_B:SO_2\)