\(n_{SO_3}=\dfrac{8}{80}=0.1\left(mol\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(0.1......................0.1\)

\(C_{M_{H_2SO_4}}=\dfrac{0.1}{0.5}=0.2\left(M\right)\)

\(n_{CuO}=\dfrac{10}{80}=0.125\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(0.1...........0.1\)

\(m_{CuO\left(dư\right)}=\left(0.125-0.1\right)\cdot80=2\left(g\right)\)

a)

\(SO_3 + H_2O \to H_2SO_4\)

Theo PTHH : \(n_{H_2SO_4} = n_{SO_2} = \dfrac{8}{80} = 0,1(mol)\)

\(\Rightarrow C\%_{H_2SO_4} = \dfrac{0,1.98}{200}.100\% = 4,9\%\)

b)

\(2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O\)

Theo PTHH : \(n_{NaOH} = 2n_{H_2SO_4} = 0,2(mol)\\ \Rightarrow m_{dd\ NaOH} = \dfrac{0,2.40}{4\%} = 200(gam)\)

a. PTHH: 2Al(OH)₃ + 3H₂SO₄ ---> Al₂(SO₄)₃ + 6H₂O

b. Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)

Ta thấy: \(\dfrac{0,75}{2}>\dfrac{0,5}{3}\)

Vậy \(Al\left(OH\right)_3\) dư.

\(m_{dư}=0,75.78-98.0,5=9,5\left(g\right)\)

c. Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,5=\dfrac{1}{6}\left(mol\right)\)

=> \(m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{6}.342=57\left(g\right)\)

a, \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right);n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PTHH: 2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

Mol:        \(\dfrac{1}{3}\)               0,5               \(\dfrac{1}{6}\)

b, Ta có: \(\dfrac{0,75}{2}>\dfrac{0,5}{3}\) ⇒ Al(OH)₃ dư, H₂SO₄ hết

⇒ \(m_{Al\left(OH\right)_3}=\left(0,75-\dfrac{1}{3}\right).78=32,5\left(g\right)\)

c, \(m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{6}.342=57\left(g\right)\)

Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{58,5}{78}=0,75\left(mol\right)\)

a. PTHH: 2Al(OH)₃ + 3H₂SO₄ ---> Al₂(SO₄)₃ + 6H₂O

b. Không có chất dư (hoặc có thể bn cho sai 49(g) dung dịch là 49(g) H₂SO₄)

c. Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.n_{Al\left(OH\right)_3}=\dfrac{1}{2}.0,75=0,375\left(mol\right)\)

=> \(m_{Al_2\left(SO_4\right)_3}=0,375.342=128,25\left(g\right)\)

\(n_{NaOH}=\dfrac{200.5\%}{100\%.40}=0,25(mol)\\ n_{HCl}=\dfrac{36,5.20\%}{100\%.36,5}=0,2(mol)\\ a,NaOH+HCl\to NaCl+H_2O\)

Vì \(\dfrac{n_{NaOH}}{1}>\dfrac{n_{HCl}}{1}\) nên \(NaOH\) dư

\(b,n_{NaOH(dư)}=0,25-0,2=0,05(mol);n_{NaCl}=0,2(mol)\\ \Rightarrow m_{\text{dd sau p/ứ}}=0,2.58,5+0,05.40=13,7(g)\\ c,m_{NaCl}=0,2.58,5=11,7(g)\\ d,n_{H_2}=0,2(mol)\\ \Rightarrow \begin{cases} C\%_{NaOH}=\dfrac{0,05.40}{36,5+200-0,2.2}.100\%=0,85\%\\ C\%_{NaCl}=\dfrac{11,7}{36,5+200-0,2.2}.100\%=4,96\% \end{cases}\)

a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b) Ta có: \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\C\%_{HCl}=\dfrac{0,4\cdot36,5}{200}\cdot100\%=7,3\%\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{MgO}+m_{ddHCl}=208\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{19}{208}\cdot100\%\approx9,13\%\)

c) PTHH: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)

Ta có: \(\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{NaOH}=\dfrac{200\cdot4\%}{40}=0,2\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) \(\Rightarrow\) NaOH p/ứ hết, MgCl₂ còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,2\left(mol\right)\\n_{Mg\left(OH\right)_2}=0,1\left(mol\right)=n_{MgCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{MgCl_2\left(dư\right)}=9,5\left(g\right)\\m_{Mg\left(OH\right)_2}=0,1\cdot58=5,8\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{ddA}+m_{ddNaOH}-m_{Mg\left(OH\right)_2}=402,2\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{402,2}\cdot100\%\approx2,91\%\\C\%_{MgCl_2\left(dư\right)}=\dfrac{9,5}{402,2}\cdot100\%\approx2,36\%\end{matrix}\right.\) 

Theo gt ta có: $n_{MgO}=0,2(mol)$

a, $MgO+2HCl\rightarrow MgCl_2+H_2O$

b, Ta có: $n_{HCl}=0,4(mol)\Rightarrow x=7,3$

Bảo toàn khối lượng ta có: $m_{ddA}=208(g)$

$\Rightarrow \%C_{MgCl_2}=9,13\%$

c, Ta có: $n_{NaOH}=0,2(mol)$

$\Rightarrow n_{Mg(OH)_2}=0,1(mol)$

Bảo toàn khối lượng ta có: $m_{ddB}=208+200-0,1.58=402,2(g)$

$\Rightarrow \%C_{MgCl_2}=2,36\%$

a)

$MgO + 2HCl \to MgCl_2  + H_2O$

b)

$n_{MgO} = \dfrac{8}{40} = 0,2(mol)$

$n_{HCl} = 2n_{MgO} = 0,4(mol)$

$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$

(x = 7,3)

$m_{dd\ A} = 8 + 200 = 208(gam)$
$C\%_{MgCl_2} = \dfrac{0,2.95}{208}.100\% = 9,13\%$

\(a.HCl+NaOH\rightarrow NaCl+H_2O\)

PỨ trung hoà 

\(b,n_{NaOH}=0,1.1=0,1mol\\ n_{NaCl}=n_{NaOH}=n_{HCl}0,1mol\\ m=m_{HCl}=0,1.36,5=3,65g\\ c,m_{NaCl}=0,1.58,5=5,85g\\ d,n_{HCl}=\dfrac{73.10}{100.36,5}=0,2mol\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,2}{1}\Rightarrow HCl.dư\\ n_{HCl,pứ}=n_{NaOH}=0,1mol\\ m_{HCl,dư}=\left(0,2-0,1\right).36,5=3,65g\)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ LTL:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ n_{HCl\left(pứ\right)}=2n_{Zn}=0,4\left(mol\right)\\\Rightarrow m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65\left(g\right)\\ b.n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ c.n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4,=4,48\left(l\right)\\ d.3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O \\ n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12\left(mol\right)\\ LTL:\dfrac{0,2}{3}< \dfrac{0,12}{1}\Rightarrow Fe_2O_3dưsauphảnứng\\ \Rightarrow n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\\ \Rightarrow m_{Fe}=\dfrac{2}{15}.56=7,467\left(g\right)\)

a) n\(Zn\)=\(\dfrac{m}{M}\)=\(\dfrac{13}{65}\)=0,2(mol)

n\(HCl\)=\(\dfrac{m}{M}\)=\(\dfrac{18,25}{36,5}=\)0,5(mol)

PTHH : Zn + 2HCl->ZnCl\(2\) + H\(2\)

            0,2     0,5

Lập tỉ lệ mol : \(^{\dfrac{0,2}{1}}\)<\(\dfrac{0,5}{2}\)

n\(Zn\) hết , n\(HCl\) dư

-->Tính theo số mol hết

               Zn + 2HCl->ZnCl\(2\) + H\(2\)

              0,2 ->  0,4      0,2        0,2

n\(HCl\) dư= n\(HCl\)(đề) - n\(HCl\)(pt)= 0,5 - 0,4 = 0,1(mol)

m​\(HCl\) dư= 0,1.36,5 = 3,65(g)

b) m\(ZnCl2\) = n.M= 0,2.136= 27,2 (g)

c)V\(H2\)=n.22,4=0,2.22,4=4,48(l)

d) n\(Fe\)\(2\)O\(3\)=\(\dfrac{m}{M}\)=\(\dfrac{19,2}{160}\)=0,12 (mol)

 3H2 +Fe2O3 → 2Fe + 3H2O

  0,2       0,12

Lập tỉ lệ mol: \(\dfrac{0,2}{3}\)<\(\dfrac{0,12}{1}\)

    nH2 hết .Tính theo số mol hết

\(HCl\)

​ 3H2 +Fe2O3 → 2Fe + 3H2O

0,2->                  0,2

m\(Fe\)=n.M= 0,2.56= 11,2(g)

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\\ 2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\)

Cho hỗn hợp tác dụng với NaOH, chất rắn không tan là Fe

=> m_Fe= 1,12 (g) \(\Rightarrow n_{Fe}=0,02\left(mol\right)\)

Ta có: \(n_{H_2\left(2\right)}=n_{Fe}=0,02\left(mol\right)\)

=> \(n_{H_2\left(1\right)}=\Sigma n_{H_2}-n_{H_2\left(2\right)}=0,065-0,02=0,045\left(mol\right)\)

\(\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2\left(1\right)}=0,03\left(mol\right)\)

\(\Rightarrow m_{Al}=0,03.27=0,81\left(g\right)\)

\(\Rightarrow\%m_{Al}=41,97\%,\%m_{Fe}=58,03\%\)

b) \(m_{FeCl_2}=0,02.127=2,54\left(g\right)\\ m_{AlCl_3}=0,03.133,5=4,005\left(g\right)\)