Bài 1

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ b,n_{Mg}=0,1(mol)\Rightarrow m_{Mg}=0,1.24=2,4(g)\\ \Rightarrow m_{MgO}=4,4-2,4=2(g)\\ c,n_{MgO}=\dfrac{2}{40}=0,05(mol)\\ \Rightarrow \Sigma n_{HCl}=0,1.2+0,05.2=0,3(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,3.36,5}{7,3\%}=150(g)\)

Bài 2:

\(n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ a,Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{H_2}=0,2(mol)\Rightarrow V_{H_2}=0,2.22,4=4,48(l)\\ b,n_{H_2SO_4}=0,2(mol)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{0,2.98}{200}.100\%=9,8\%\\ c,n_{FeSO_4}=0,2(mol)\\ \Rightarrow C\%_{FeSO_4}=\dfrac{0,2.152}{11,2+200-0,2.2}.100\%=14,42\%\)

30 cn trồng trong \(45\cdot14:30=21\left(ngày\right)\)

\(a,2Cu+O_2\xrightarrow{t^o}2CuO\\ CuO+2HCl\to CuCl_2+H_2\\ CuCl_2+2AgNO_3\to 2AgCl\downarrow+Cu(NO_3)_2\\ Cu(NO_3)_2+2NaOH\to Cu(OH)_2\downarrow+2NaNO_3\\ Cu(OH)_2+H_2SO_4\to CuSO_4+2H_2O\)

\(b,Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ Al_2(SO_4)_3+3BaCl_2\to 2AlCl_3+3BaSO_4\downarrow\\ AlCl_3+3NaOH\to Al(OH)_3\downarrow+3NaCl\\ 2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ c,CaO+CO_2\to CaCO_3\\ CaCO_3+2HCl\to CaCl_2+H_2O+CO_2\uparrow\\ CaCl_2+2AgNO_3\to 2AgCl\downarrow+Ca(NO_3)_2\)

\(d,2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ FeCl_3+3AgNO_3\to 3AgCl\downarrow+Fe(NO_3)_3\\ 3Zn+2Fe(NO_3)_3\to 2Fe+3Zn(NO_3)_2\\ Fe+2HCl\to FeCl_2+H_2\\ FeCl_2+2NaOH\to Fe(OH)_2\downarrow+2NaCl\\ Fe(OH)_2+H_2SO_4\to FeSO_4+2H_2O\)

\(n_{SO_2}=\dfrac{4,958}{24,79}=0,2(mol)\\ \Rightarrow m_{SO_2}=0,2.64=12,8(g)\)

\(n_{SO_2}=\dfrac{4,958}{24,79}=0,2(mol)\)

Câu 5 (tiếp)

\(21,Al(OH)_3+3HCl\to AlCl_3+3H_2O\\ 22,2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ 23,Mg(OH)_2+H_2SO_4\to MgSO_4+2H_2O\\ 24,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ 25,KOH+HNO_3\to KNO_3+H_2O\\ 26,Cu(OH)_2+2HNO_3\to Cu(NO_3)_2+2H_2O\\ 27,Al(OH)_3+3HNO_3\to Al(NO_3)_3+3H_2O\\ 28,3NaOH+H_3PO_4\to Na_3PO_4+3H_2O\\ 29,3Fe(OH)_2+2H_3PO_4\to Fe_3(PO_4)_2+6H_2O\\ 30,2Fe(OH)_3+2H_3PO_4\to 2FePO_4+6H_2O\)

Câu 5:

\(1,2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ 2,4P+5O_2\xrightarrow{t^o}2P_2O_5\\ 3,Zn+H_2SO_4\to ZnSO_4+H_2\\ 4,2Al+6HCl\to 2AlCl_3+3H_2\\ 5,2KClO_3\xrightarrow{t^o}2KCl+3O_2\\ 6,2Mg+O_2\xrightarrow{t^o}2MgO\\ 7,Na_2O+2HCl\to 2NaCl+H_2O\\ 8,CaO+2HCl\to CaCl_2+H_2\\ 9,Al_2O_3+6HCl\to AlCl_3+3H_2O\\ 10,Ag_2O+2HNO_3\to 2AgNO_3+H_2O\\ 11,MgO+2HNO_3\to Mg(NO_3)_2+H_2O\\ 12,Fe_2O_3+6HNO_3\to 2Fe(NO_3)_3+3H_2O\)

\(13,K_2O+H_2SO_4\to K_2SO_4+H_2O\\ 14,ZnO+H_2SO_4\to ZnSO_4+H_2O\\ 15,Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ 16,3Na_2O+2H_3PO_4\to 2Na_3PO_4+3H_2O\\ 17,3BaO+2H_3PO_4\to Ba_3(PO_4)_2+3H_2O\\ 18,Fe_2O_3+2H_3PO_4\to 2FePO_4+3H_2O\\ 19,KOH+HCl\to KCl+H_2O\\ 20,Ba(OH)_2+2HCl\to BaCl_2+2H_2O\)

\(n_{Zn}=\dfrac{9,75}{65}=0,15(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{H_2}=0,15(mol);n_{HCl}=0,3(mol)\\ a,V_{H_2}=0,15.22,4=3,36(l)\\ b,m_{HCl}=0,3.36,5=10,95(g)\)

Chọn C

Do NaOH là dd Bazơ nên làm quỳ hóa xanh

Còn Na₂SO₄ và NaCl là muối, HCl là dd Axit

Trong 1 mol hợp chất

\(\left\{{}\begin{matrix}n_{Ba}=\dfrac{208\cdot65,87\%}{137}\approx1\left(mol\right)\\n_{Cl_2}=\dfrac{208-1\cdot137}{71}=1\left(mol\right)\end{matrix}\right.\\ \Rightarrow CTHH:BaCl_2\)