1.\(6KOH+3Al_2\left(SO_4\right)_3\rightarrow3K_2SO_4+2Al\left(OH\right)_3\\ 2.Fe_xO_y+yCO\rightarrow xFe+CO_2\\ 4FeS_2+11O_2\rightarrow2Fe_2O_3+8SO_2\\ 8Al+30HNO_3\rightarrow8Al\left(NO_3\right)_3+3N_2O+15H_2O\)
Câu 2:
Đặt \(n_{CaO}=x(mol);n_{MgO}=y(mol)\)
\(\Rightarrow 56x+40y=7,6(1)\\ PTHH:CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow\\ MgCO_3\xrightarrow{t^o}MgO+CO_2\uparrow\\ CO_2+2NaOH\to Na_2CO_3+H_2O\\ n_{Na_2CO_3}=\dfrac{15,9}{106}=0,15(mol)\\ \Rightarrow n_{CO_2}=0,15(mol)\\ \Rightarrow x+y=0,15(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,05(mol)\\ \Rightarrow n_{CaCO_3}=0,1(mol);n_{MgCO_3}=0,05(mol)\\ \Rightarrow m_{hh}=m_{CaCO_3}+m_{MgCO_3}=0,1.100+0,05.84=14,2(g)\)
Câu 3:
Đặt hóa trị M là x(mol)
\(PTHH:2M+2xHCl\to 2MCl_x+xH_2\\ \Rightarrow x.n_{M}=2n_{H_2}=\dfrac{5,6.2}{22,4}=0,5(mol)\\ \Rightarrow \dfrac{x.16,25}{M_{M}}=0,5\\ \Rightarrow M_{M}=x.32,5\)
Thay \(x=2\Rightarrow M_M=65(g/mol)\)
Vậy M là kẽm(Zn)
\(b,n_{HCl}=2n_{H_2}=0,5(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,5}{0,2}=2,5(l)\)
Câu 2:
\(a,\%_{Cu}=\dfrac{64}{160+5.18}.100\%=25,6\%\\ \Rightarrow m_{Cu}=140,625.25,6\%=36(g)\)
Câu 1 bạn TK bài chị Thảo Phương, mình cân bằng câu b sai nha
