\(2X+2nHCl \to 2XCl_n+nH_2\)

\(a/ CuO+2HCl \to CuCl_2+H_2O\\ b/\\ n_{CuO}=0,125(mol)\\ \to n_{HCl}=0,125.2=0,25(mol)\\ m_{HCl}=0,25.36,5=9,125(g)\\ c/\\ n_{CuO}=n_{CuCl_2}=0,125(mol)\\ CM_{CuCl_2}=\frac{0,125}{0,5}=0,25M\)

\(a/\\ Tổng: 2p+n=40(1)\\ MĐ > KMĐ: 2p-n=12(2)\\ (1)(2)\\ p=e=13\\ n=14\\ A=13+14=27\\ b/\\ Che: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\)

\(Mg+2HCl \to MgCl_2+H_2\\ n_{H_2}=0,15(mol)\\ \to n_{Mg}=n_{H_2}=0,15(mol)\\ \%m_{Mg}=\frac{0,15.24}{10}.100\%=36\%\\ \%m_{Cu}=100\%-36\%=64\%\)

\(2X+nCl_2 \xrightarrow{t^{o}} 2XCl_n\\ n_{X}=a(mol)\\ n_{XCl_n}=a(mol)\\ \frac{a.X}{a.(X+35,5n)}=\frac{1,625}{3,4}\\ \to \frac{X}{X+35,5n}=\frac{1,625}{3,4}\\ \to X=32,5n(g/mol)\\ a=2; X=65 (Zn) \)

\(2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2\\ n_{Al}=0,4(mol)\\ a/\\ n_{H_2}=\frac{3}{2}.0,4=0,6(mol)\\ V_{H_2}=0,6.22,4=13,44(l)\\ b/\\ n_{H_2SO_4}=n_{H_2}=0,6(mol)\\ n_{ddH_2SO_4}=\frac{0,6.98.100}{20}=294(g)\\ c/\\ n_{Al_2(SO_4)_3}=0,2(mol)\\ C\%_{Al_2(SO_4)_3}=\frac{0,2.342}{10,8+294-0,6.2}.100\%=22,52\%\)

\((1) Zn+H_2SO_4 \to ZnSO_4+H_2\\ (2) ZnSO_4+BaCl_2 \to BaSO_4+ZnCl_2\\ (3) ZnCl_2+2AgNO_3 \to 2AgCl+Zn(NO_3)_2\\ (4) Zn(NO_3)_2+2NaOH \to Zn(OH)_2+2NaNO_3\\ (5) Zn(OH)_2 \xrightarrow{t^{o}} ZnO+H_2O\)

:33 xin nhá :33

\(\text{Đặt công thức chung là: R}\\ R+2HCl \to RCl_2+H_2\\ n_{H_2}=0,05(mol)\\ n_R=n_{H_2}=0,1(mol)\\ M_R=\frac{3,1}{0,1}=31(g/mol)\\ M_Na=23 \leq M_R=31 \leq M_{K}=39\\ \to B\)

WOW bạn giỏi thiệc ó :3333