Câu 1:

$a,Mg+2HCl\to MgCl_2+H_2\\ b,2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ c,2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ d,4Fe_xO_y+(3x-2y)O_2\to 2xFe_2O_3$

Câu 2:

$a,CTTQ:Cu_x^{II}Cl_y^I\\ \text{Theo QTHT, ta có: }x.II=y.I\Rightarrow \dfrac{x}{y}=\dfrac{1}{2}\\ \Rightarrow x=1;y=2\\ \Rightarrow CTHH:CuCl_2\\ PTK_{CuCl_2}=64+35,5.2=135(đvC)\\ CTTQ:Mg_x^{II}(PO_4)_y^{III}\\ \text{Theo QTHT, ta có: }x.II=y.III\Rightarrow \dfrac{x}{y}=\dfrac{3}{2}\\ \Rightarrow x=3;y=2\\ \Rightarrow CTHH:Mg_3(PO_4)_2\\ PTK_{Mg_3(PO_4)_2}=24.3+(31+16.4).2=262(đvC)$

Câu 3:

$1,M_{Al(OH)_3}=27+(16+1).3=78(g/mol)\\ \%m_{Al}=\dfrac{27}{78}.100\%=34,62\%\\ \%m_O=\dfrac{16.3}{78}.100\%=61,54\%\\ \%m_H=100\%-61,54\%-34,62\%=3,84\%\\ 2,\\ a,PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ \text{Tỉ lệ: }3:2:1\\ b,n_{Fe}=\dfrac{25,2}{56}=0,45(mol)\\ \Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=0,3(mol);n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,15(mol)\\ \Rightarrow V_{O_2}=0,3.22,4=6,72(l)\\ m_{Fe_3O_4}=0,15.232=34,8(g)$

C1 a,Mg+2HCl→MgCl2+H2 b,2Fe+3Cl2to→2FeCl3 c,2NaOH+H2SO4→Na2SO4+2H2O d,2FexOy+(3x−2y)/2O2→xFe2O3 C2 a, Gọi CtCt là CuxCly. Vì Cu hóa trị ll và Cl hóa trị l ADQTHT ta có x.ll=y.l =>x=1;y=2 =>CuCl2 b, Gọi CTCT là Mgx(PO4)y Vì Mg hóa trị ll và PO4 hóa trị lll ADQTHT ta có x.ll=y.lll =>x=3;y=2 =>Mg3(PO4)2 Câu 3 MAl2(OH)2=27×2+(16+1)×2=88g/mol Trong 1mol hc Al2(OH)2 có: 2mol ntử Al; 2mol ntử O; 2mol ntử H +,%mAl=27×2×100÷88=61,26% +,%mO=16×2×100÷88=27,27% +,%mH=100-61,26-27,27=11,47%