a) $n_{NaOH} = \dfrac{8}{40} = 0,2(mol)$
c) $n_{CO_2} = \dfrac{6,72}{22,4} = 0,3(mol)$

a) \(n_{HCl}=\dfrac{73}{36,5}=2\left(mol\right)\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

            1<-----1------------------->1

=> m = 65.1 = 65 (g)

b) PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)

                 \(\dfrac{1}{3}\)<-------1

=> \(m_{Fe_2O_3}=\dfrac{1}{3}.160=\dfrac{160}{3}\left(g\right)\)

a) $2Al +3 H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b) $n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{0,4}{3}(mol)$
$m_{Al} = \dfrac{0,4}{3}.27 = 3,6(gam)$

c) $2H_2 + O_2 \xrightarrow{t^o} 2H_2O$

Theo PTHH : $V_{O_2} = \dfrac{1}{2}V_{H_2} = 2,24(lít)$
$V_{không\ khí} = V_{O_2} : 21\% = 2,24 : 21\% = 10,67(lít)$

a) PTHH: `2Al + 3H_2SO_4 -> Al_2(SO_4)_3`

b) $n_{H_2} = \dfrac{4,48}{22,4} = 0,2 (mol)$

Theo PTHH: $n_{Al} = \dfrac{2}{3} n_{H_2} = \dfrac{2}{15} (mol)$

`=> m_{Al} = 2/(15) . 27 = 3,6 (g)`

c) PTHH: $2H_2 + O_2 \xrightarrow{t^o} 2H_2O$

Theo PTHH: $n_{O_2} = \dfrac{1}{2} n_{H_2} = 0,1 (mol)$

$\rightarrow V_{kk} = \dfrac{0,1.22,4.100}{21} = \dfrac{64}{3} (l)$

a) $Mg + 2HCl \to MgCl_2 + H_2$

b) Theo PTHH : $n_{Mg} = n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)$
$m_{Mg} = 0,3.24 = 7,2(gam)$

c) $n_{MgCl_2} = n_{H_2} = 0,3(mol)$
$m_{MgCl_2} = 0,3.95 = 28,5(gam)$

d) $n_{Fe_2O_3} = \dfrac{24}{160} = 0,15(mol)$
$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$
Ta thấy : 

$n_{Fe_2O_3} : 1 > n_{H_2} : 3$ nên $Fe_2O_3$ dư

$n_{Fe_2O_3\ pư} = \dfrac{1}{3}n_{H_2} = 0,1(mol)$
$m_{Fe_2O_3\ dư} = 24 - 0,1.160 = 8(gam)$

a) PTHH: Mg + 2HCl ---> MgCl₂ + H₂ 

b) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PTHH: \(n_{Mg}=n_{MgCl_2}=n_{H_2}=0,3\left(mol\right)\)

=> m_Mg = 0,3.24 = 7,2 (g)

c) m_muối = 0,3.95 = 28,5 (g)

d) \(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\)

PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,3}{3}\) => Fe₂O₃ dư, H₂ hết

Theo PTHH: \(n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\)

=> m_{Fe2O3 dư} = (0,15 - 0,1).160 = 8 (g)

a) $Fe + 2HCl \to FeCl_2 + H_2$

Theo PTHH : $n_{H_2} = n_{Fe} = \dfrac{8,4}{56} = 0,15(mol)$
$V_{H_2} = 0,15.22,4 = 3,36(lít)$

b) $n_{FeCl_2} = n_{Fe} = 0,15(mol)$

$m_{FeCl_2} = 0,15.127 = 19,05(gam)$

c) $n_{CuO} = \dfrac{8}{80} = 0,1(mol)$

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

$n_{CuO} : 1 < n_{H_2} : 1$ nên $H_2$ dư

$n_{Cu} = n_{CuO} = 0,1(mol)$
$m_{Cu} = 0,1.64 = 6,4(gam)$

a) \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

           0,15-------------->0,15---->0,15

=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

b) \(m_{muối}=0,15.127=19,05\left(g\right)\)

c) \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\) => CuO phản ứng hết, H₂ còn dư

Theo PTHH: \(n_{Cu}=n_{CuO}=0,1\left(mol\right)\)

=> m_Cu = 0,1.64 = 6,4 (g)

Zn+2HCl->ZnCl2+H2

0,5------------------0,5

H2+FeO-to>Fe+H2O

0,5---------------0,5

b)n Zn=0,5 mol

=>VH2=0,5.22,4=11,2l

c)n FeO=0,7 mol

=>FeO dư

=>m Fe=0,5.56=28g

a) $m_{dd} = \dfrac{m_{đường}}{C\%} = \dfrac{50}{25\%} = 200(gam)$
b) $m_{nước} = m_{dd} - m_{đường} = 200 - 50 = 150(gam)$

$C\%_{muối\ ăn} = \dfrac{S}{S + 100}.100\% = \dfrac{36}{36 + 100}.100\% = 26,47\%$

$C\%_{đường} = \dfrac{S}{S + 100}.100\% = \dfrac{204}{204 + 100}.100\% = 67,1\%$

Độ tan của một chất trong nước là số gam chất tan tan tối đa trong 100 gam dung môi.

$S_{Na_2CO_3} = \dfrac{53}{250}.100 = 21,2(gam)$

$2Na + 2H_2O \to 2NaOH + H_2$
$Ca + 2H_2O \to Ca(OH)_2 + H_2$
$Ba + 2H_2O \to Ba(OH)_2 + H_2$
$K_2O + H_2O \to 2KOH$
$SO_3 + H_2O \to H_2SO_4$
$CO_2 + H_2O \rightleftharpoons H_2CO_3$
$P_2O_5 + 3H_2O \to 2H_3PO_4$

$BaO + H_2O \to Ba(OH)_2$
$CaO + H_2O \to Ca(OH)_2$
$Na_2O + H_2O \to 2NaOH$
$N_2O_5 + H_2O \to 2HNO_3$