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Câu 5:
\(1,2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ 2,4P+5O_2\xrightarrow{t^o}2P_2O_5\\ 3,Zn+H_2SO_4\to ZnSO_4+H_2\\ 4,2Al+6HCl\to 2AlCl_3+3H_2\\ 5,2KClO_3\xrightarrow{t^o}2KCl+3O_2\\ 6,2Mg+O_2\xrightarrow{t^o}2MgO\\ 7,Na_2O+2HCl\to 2NaCl+H_2O\\ 8,CaO+2HCl\to CaCl_2+H_2\\ 9,Al_2O_3+6HCl\to AlCl_3+3H_2O\\ 10,Ag_2O+2HNO_3\to 2AgNO_3+H_2O\\ 11,MgO+2HNO_3\to Mg(NO_3)_2+H_2O\\ 12,Fe_2O_3+6HNO_3\to 2Fe(NO_3)_3+3H_2O\)
\(13,K_2O+H_2SO_4\to K_2SO_4+H_2O\\ 14,ZnO+H_2SO_4\to ZnSO_4+H_2O\\ 15,Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ 16,3Na_2O+2H_3PO_4\to 2Na_3PO_4+3H_2O\\ 17,3BaO+2H_3PO_4\to Ba_3(PO_4)_2+3H_2O\\ 18,Fe_2O_3+2H_3PO_4\to 2FePO_4+3H_2O\\ 19,KOH+HCl\to KCl+H_2O\\ 20,Ba(OH)_2+2HCl\to BaCl_2+2H_2O\)
Câu 5 (tiếp)
\(21,Al(OH)_3+3HCl\to AlCl_3+3H_2O\\ 22,2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ 23,Mg(OH)_2+H_2SO_4\to MgSO_4+2H_2O\\ 24,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ 25,KOH+HNO_3\to KNO_3+H_2O\\ 26,Cu(OH)_2+2HNO_3\to Cu(NO_3)_2+2H_2O\\ 27,Al(OH)_3+3HNO_3\to Al(NO_3)_3+3H_2O\\ 28,3NaOH+H_3PO_4\to Na_3PO_4+3H_2O\\ 29,3Fe(OH)_2+2H_3PO_4\to Fe_3(PO_4)_2+6H_2O\\ 30,2Fe(OH)_3+2H_3PO_4\to 2FePO_4+6H_2O\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mg + HCl \(\rightarrow\)MgCl2 + H2 (1)
Fe + HCl \(\rightarrow\)FeCl2 + H2 (2)
mHCl=\(300.\dfrac{7,3}{100}=21,9\left(g\right)\)
nHCl=\(\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
Đặt nMg=a
nFe=b
Ta có:
\(\left\{{}\begin{matrix}24a+56b=13,6\\2a+2b=0,6\end{matrix}\right.\)
a=0,1;b=0,2
mMg=24.0,1=2,4(g)
mFe=13,6-2,4=11,2(g)
C% Mg=\(\dfrac{2,4}{13,6}.100\%=17,6\%\)
C% Fe=100-17,6=82,4%
c;Theo PTHH 1 ta có:
nMgCl2=nMg=0,1(mol)
mMgCl2=0,1.95=9,5(g)
Theo PTHH 2 ta có:
nFeCl2=nFe=0,2(mol)
mFeCl2=0,2.127=25,4(g)
C% dd MgCl2=\(\dfrac{9,5}{300+13,6-0,6}.100\%=3\%\)
C% dd FeCl2=\(\dfrac{25,4}{300+13,6-0,6}.100\%=8\%\)
d;
d;CuO + H2\(\rightarrow\)Cu + H2O ( 3)
nH2=0,2+0,1=0,3(mol)
nCuO=\(\dfrac{40}{80}=0,5\left(mol\right)\)
Vậy CuO dư 0,2 mol
Theo PTHH 3 ta có:
nH2=nCu=0,3(mol)
mCu=0,3.64=19,2(g)
mCuO còn lại=0,2.80=16(g)
mchất rắn=16+19,2=35,2(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Fe + 2HCl --> FeCl2 + H2
b) Fe + 2HCl --> FeCl2 + H2
__0,5-->1
=> mHCl = 1.36,5 = 36,5 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 18: C
Câu 19: D
Câu 20: B
Câu 21: C
Câu 22: B
Câu 23: C