\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow\left(trắng\right)+H_2O\\ CuSO_4+BaCl_2\rightarrow BaSO_4\downarrow\left(trắng\right)+CuCl_2\)

\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\\ a,PTHH:2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\\ b,n_{Na_2SO_4}=n_{H_2}=n_{H_2SO_4}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ m_{Na_2SO_4}=142.0,05=7,1\left(g\right)\\ c,C\%_{ddH_2SO_4}=\dfrac{0,05.98}{4,9}.100\%=100\%\)

Thường C% < 100% ớ em

\(V_{rượu\left(ng.chất\right)}=120.0,4+x.0,1=48+0,1x\left(ml\right)\\ V_{hh\left(rượu,nước\right)}=120+x\left(ml\right)\\ D_{r\left(sau\right)}=20^o\\ \Leftrightarrow\dfrac{48+0,1x}{120+x}.100\%=20\%\\ \Leftrightarrow x=240\left(ml\right)\)

Vậy cần trộn 120ml rượu 40 độ với 240ml rượu 10 độ để thu được dd rượu 20 độ

ⁿH₂ = 6,72 : 22,4 = 0,3 ( mol ) 

PTHH :    2Al +  6HCl    ->     2AlCl₃   + 3H₂ 

                0,2     0,6 mol <-    0,3mol

a = ^mAl = 0,2 x 27 = 5,4 (g)

^VHCl = 0,6 : 2 = 0,3 ( l ) = 300 ( ml )

\(a,PTHH:Fe_2O_3+3H_2\rightarrow\left(t^o\right)3H_2O+2Fe\\ n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\ b,n_{Fe}=2.0,1=0,2\left(mol\right)\\ m_{Fe}=0,2.56=11,2\left(g\right)\)

\(C1\\ n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,H_2+CuO\rightarrow\left(t^o\right)Cu+H_2O\\ n_{Cu\left(LT\right)}=n_{H_2}=0,3\left(mol\right)\\ n_{Cu\left(TT\right)}=n_{Cu\left(LT\right)}.H=0,3.80\%=0,24\left(mol\right)\\ m_{Cu\left(TT\right)}=0,24.80=19,2\left(g\right)\)

- Oxit:

+ Oxit bazo: Al₂O₃ (Nhôm oxit), CuO (Đồng (II) oxit)

+ Oxit axit: N₂O₅ (dinito pentaoxit)

- Axit: 

H₂CO₃ (Axit cacbonic), H₃PO₄ (axit photphoric)

- Bazo:

KOH (Kali hidroxit), Zn(OH)₂ (Kẽm hidroxit)

- Muối: 

KHSO₄ (Kali hidrosunfat), CuCl₂ (Đồng (II) clorua), ZnSO₄ (Kẽm sunfat)

a)

$4K + O_2 \xrightarrow{t^o} 2K_2O$
$2Ca + O_2 \xrightarrow{t^o} 2CaO$
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
$C + O_2 \xrightarrow{t^o} CO_2$

$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
b)

$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe +3 H_2O$
$PbO + H_2 \xrightarrow{t^o} Pb + H_2O$

c)

$Ca + 2H_2O \to Ca(OH)_2 + H_2$
$2K + 2H_2O \to 2KOH + H_2$
$CaO + H_2O \to Ca(OH)_2$

$M_{CH_4O} = 12 + 4 + 16 = 32$

$M_{C_2H_5ONa} = 12.2 + 5 + 16 + 23 = 68$
$M_{CH_3Cl} = 12 + 3 + 35,5 = 50,5$
$M_{C_2H_4} = 12.2 + 4 = 28$

Chọn đáp án B