\(n_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\) 

  0,02             0,04         0,02     0,02                ( mol )

\(C_{M_{HCl}}=\dfrac{0,04}{0,02}=2\left(M\right)\)

\(m_{Na_2CO_3}=0,02.106=2,12\left(g\right)\)

\(m_{NaCl\left(hh\right)}=5-2,12=2,88\left(g\right)\)

\(m_{muối}=2,88+\left(0,02.58,5\right)=4,05\left(g\right)\)

`c.\(\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{2,12}{5}.100=42,4\%\\\%m_{NaCl}=100\%-42,4\%=57,6\%\end{matrix}\right.\)

\(a,Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ n_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ n_{Na_2CO_3}=n_{CO_2}=0,02\left(mol\right)\\ n_{HCl}=0,02.2=0,04\left(mol\right)\\ C_{MddHCl}=\dfrac{0,04}{0,02}=2\left(M\right)\\ b,m_{NaCl\left(sau\right)}=m_{NaCl\left(bđ\right)}+m_{NaCl\left(sp\right)}=\left(5-0,02.106\right)+0,02.2.58,5=5,22\left(g\right)\\ c,\%m_{Na_2CO_3}=\dfrac{0,02.106}{5}.100\%=42,4\%\\ \Rightarrow\%m_{NaCl}=100\%-42,4\%=57,6\%\)

a)

$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
b) $n_P = \dfrac{3,72}{31} = 0,12(mol) ; n_{O_2} = \dfrac{4,032}{22,4} = 0,18(mol)$

Ta thấy : 

$n_P : 4 < n_{O_2} : 5$ nên $O_2$ dư

$n_{P_2O_5} = \dfrac{1}{2}n_P = 0,06(mol)$
$m_{P_2O_5} = 0,06.142 = 8,52(gam)$

c) Hiện tượng : quỳ tím hoá đỏ

$P_2O_5 + 3H_2O \to 2H_3PO_4$

$n_{H_3PO_4} = 2n_{P_2O_5} = 0,12(mol)$
$C_{M_{H_3PO_4}} = \dfrac{0,12}{0,3} = 0,4M$

`a.b.`

\(n_P=\dfrac{3,72}{31}=0,12\left(mol\right)\) ; \(n_{O_2}=\dfrac{4,032}{32}=0,126\left(mol\right)\)

\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

0,12 >  0,126                   ( mol )

             0,126       0,0504     ( mol )

\(m_{P_2O_5}=0,0504.142=7,1568\left(g\right)\)

`c.`

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

0,0504                     0,1008      ( mol )

Khi nhỏ dd A lên mẩu quỳ tím thì quỳ tím sẽ chuyển sang màu đỏ

\(C_{M_{H_3PO_4}}=\dfrac{0,1008}{0,3}=0,336\left(M\right)\)

`n_[KOH]=0,15.0,1=0,015(M)`

`m_[KOH]=0,015.56=0,84(g)`

\(m_{NaCl}=30\%.80=24g\)

\(n_{NaCl}=\dfrac{24}{58,5}=0,41mol\)

bạn cần giúp bài nào ạ

`1.

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) ; \(n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\)

X là ankan: \(C_nH_{2n+2}\)

\(n_X=n_{H_2O}-n_{CO_2}=0,4-0,3=0,1\left(mol\right)\)

\(S_C=\dfrac{n_{CO_2}}{n_X}=\dfrac{0,3}{0,1}=3\)

`=>` CTHH: \(C_3H_8\)

\(n_{O_2}=n_{CO_2}+\dfrac{1}{2}n_{H_2O}=0,3+\dfrac{1}{2}.0,4=0,5\left(mol\right)\)

\(V_{O_2}=0,5.22,4=11,2\left(l\right)\)

`2.`

\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)

\(2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\)

Đặt \(\left\{{}\begin{matrix}n_{C_2H_4}=x\\n_{C_4H_{10}}=y\end{matrix}\right.\) ( mol )

\(\Rightarrow m_{hh}=28x+58y=23\left(1\right)\)

\(\overline{M}_{hhk}=23.2=46\) \((g/mol)\)

\(\Rightarrow\dfrac{28x+58y}{x+y}=46\) \(\Leftrightarrow18x-12y=0\left(2\right)\)

\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

\(V_{CO_2}=\left(2.0,2+4.0,3\right).22,4=35,84\left(l\right)\)

\(m_{H_2O}=\left(2.0,2+5.0,3\right).18=34,2\left(g\right)\)

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,2}{0,2+0,3}.100=40\%\\\%V_{C_4H_{10}}=100\%-40\%=60\%\end{matrix}\right.\)

`3.` Giống bài 2

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

0,1                                 0,1                     ( mol )

\(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)

\(m_{ddspứ}=10,2+300=310,2\left(g\right)\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{310,2}.100=11,02\%\)

a) $Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$

b) $n_{Al_2O_3} = \dfrac{10,2}{102} = 0,1(mol)$

Theo PTHH : 

$n_{H_2SO_4} = 3n_{Al_2O_3} = 0,3(mol)$

$\Rightarrow m_{dd\ sau\ pư} = 300 + 10,2 = 310,2(gam)$

$C\%_{Al_2(SO_4)_3} = \dfrac{0,1.342}{310,2}.100\% = 11\%$

a) $Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$

b) $n_{Al_2O_3} = \dfrac{10,2}{102} = 0,1(mol)$

Theo PTHH : 

$n_{H_2SO_4} = 3n_{Al_2O_3} = 0,3(mol)$

Gọi $m_{dd\ H_2SO_4} = a(gam)$

$\Rightarrow m_{dd\ sau\ pư} = a + 10,2(gam)$

$C\%_{Al_2(SO_4)_3} = \dfrac{0,1.342}{a + 10,2}.100\% = \dfrac{34,2}{a + 10,2}.100\%$

\(n_{Al_2O_3}=\dfrac{10.2}{10,2}=0,1mol\)

\(Al_2O_3+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2O\)

Chưa đủ dữ kiện tính \(C\%_{Al_2\left(SO_4\right)_3}\) vì chưa có khối lượng dd sulfuric acid.

\(a,Đặt:n_{Mg}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}24a+56b=18,4\\22,4+22,4b=11,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,3.24}{18,4}.100\approx39,13\%\\ \Rightarrow\%m_{Fe}\approx60,87\%\\ b,n_{HCl}=2.\left(a+b\right)=1\left(mol\right)\\ C_{MddHCl}=\dfrac{1}{0,5}=2\left(M\right)\)

Thanks 

\(a,CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ n_{HCl}=2.0,02=0,04\left(mol\right)\\ C_{MddHCl}=\dfrac{0,04}{0,2}=0,2\left(M\right)\\ b,n_{CaCO_3}=n_{CO_2}=0,02\left(mol\right)\\\Rightarrow \%m_{CaCO_3}=\dfrac{0,02.100}{5}.100\%=40\%\\ \Rightarrow\%m_{CaSO_4}=100\%-40\%=60\%\)

`CaCO_3 + 2HCl -> CaCl_2 + H_2 O + CO_2 \uparrow`

  `0,02`       `0,04`                                          `0,04`        `(mol)`

`CaSO_4 + 2HCl -> CaCl_2 + H_2 SO_4`

   `3/136`        `3/68`                                          `(mol)`

`n_[CO_2]=[0,448]/[22,4]=0,02(mol)`

`n_[CaSO_4]=[5-0,02.100]/[136]=3/136(mol)`

`a)C_[M_[HCl]]=[0,04+3/68]/[0,2]=0,4(M)`

`b)%m_[CaCO_3]=[0,02.100]/5 .100=40%`

    `%m_[CaSO_4]=100-40=60%`