Câu 3 : 

\(m_{ct}=\dfrac{10.80}{100}=8\left(g\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

a) Hiện tượng : Xuất hiện kết tủa trắng

Pt : \(2NaOH+MgSO_4\rightarrow Na_2SO_4+Mg\left(OH\right)_2|\)

           2                1                  1               1

        0,2                0,1               0,1            0,1

\(n_{Mg\left(OH\right)2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{Mg\left(OH\right)2}=0,1.58=5,8\left(g\right)\)

b) \(n_{MgSO4}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

 \(m_{MgSO4}=0,1.120=12\left(g\right)\)

\(m_{ddMgSO}=\dfrac{12.100}{10}=120\left(g\right)\)

c) \(n_{Na2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{Na2SO4}=0,1.142=14,2\left(g\right)\)

\(m_{ddspu}=80+120-5,8=194,2\left(g\right)\)

\(C_{Na2SO4}=\dfrac{14,2.100}{194,2}=7,31\)⁰/₀

 Chúc bạn học tốt

Câu 4 : 

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

         1         1               1           1

        0,2      0,2

\(n_{H2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{19,6.100}{20}=98\left(g\right)\)

\(V_{ddH2SO4}=\dfrac{98}{1,2}\simeq81,67\left(ml\right)\)

 Chúc bạn học tốt

\(n_{CuCl_2}=\dfrac{60,75}{135}=0,45mol\\ a)CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
0,45           0,9              0,45                     0,9
\(b)m_X=m_{Cu\left(OH\right)_2}=0,45.81=36,45g\\ c)m_{ddNaOH}=\dfrac{0,9.40}{15\%}\cdot100\%=240g\\ d)m_{ddNaCl}=60,75+240-36,45=264,3g\\ C_{\%NaCl}=\dfrac{0,9.58,5}{264,3}\cdot100\%=19,92\%\\ e)n_{H_2SO_4}=\dfrac{245.20\%}{100\%.98}=0,5mol\\ H_2SO_4+Cu\left(OH\right)_2\rightarrow CuSO_4+2H_2O\\ \Rightarrow\dfrac{0,5}{1}>\dfrac{0,45}{1}\Rightarrow H_2SO_4.dư\)
\(\Rightarrow\)Dung dịch acid \(H_2SO_4\) làm tan hết chất X\(\left(Cu\left(OH\right)_2\right)\)

$MgSO_4 + 2NaOH \to Mg(OH)_2 + Na_2SO_4$
$Mg(OH)_2 \xrightarrow{t^o} MgO + H_2O$

n NaOH = 120.16%/40 = 0,48(mol)

n Mg(OH)2 = 1/2 n NaOH = 0,24(mol)

n MgO = n Mg(OH)2 = 0,24(mol)

=> m = 0,24.40 = 9,6(gam)

\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3(mol)\\ 3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=0,3(mol);n_{NaOH}=n_{NaCl}=0,9(mol)\\ a,m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ \)

\(d,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}=193,42(ml)\)

\(n_{H_2SO_4}=\dfrac{200\cdot19.6\%}{98}=0.4\left(mol\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(0.4......................0.4\)

\(m_{SO_3}=0.4\cdot80=32\left(g\right)\)

\(b.\)

\(n_{H_2SO_4}=\dfrac{80\cdot19.6\%}{98}=0.16\left(mol\right)\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(0.16..........0.16..............0.16\)

\(m_{MgO}=0.16\cdot40=6.4\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=6.4+80=86.4\left(g\right)\)

\(C\%MgSO_4=\dfrac{0.16\cdot120}{86.4}\cdot100\%=22.22\%\)

a)

$SO_3 + H_2O \to H_2SO_4$

n SO3 = n H2SO4 = 200.19,6%/98 = 0,4(mol)

=> m = 0,4.80 = 32(gam)

b)

$MgO + H_2SO_4 \to MgSO_4 + H_2O$

n MgSO4 = n MgO = n H2SO4 = 80.19,6%/98 = 0,16(mol)

=> m MgO = 0,16.40 = 6,4(gam)

Sau pư, m dd =  6,4 + 80 = 86,4(gam)

=> C% MgSO4 = 0,16.120/86,4   .100% = 22,22%

a, \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

b, Ta có: \(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

Theo PT: \(n_{MgO}=n_{MgSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)

c, Ta có: m _{dd sau pư} = 8 + 200 = 208 (g)

\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{208}.100\%\approx11,54\%\)

a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)=n_{MgCl_2}\)

\(\Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)

c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(p.ứ\right)}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl\left(p.ứ\right)}=0,3\cdot36,5=10,95\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)  

`a)PTHH:`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

`0,2`     `0,6`                               `0,3`       `(mol)`

`n_[Al]=[5,4]/27=0,2(mol)`

`b)V_[H_2]=0,3.22,4=6,72(l)`

`c)m_[dd HCl]=[0,6.36,5]/10 . 100 =219(g)`

n_Al  = \(\dfrac{5,4}{27}\) = 0,2 (mol)

 2Al + 6HCl -> 2AlCl₃ + 3H₂

     0,2     0,6       0,2      0,3

=>VH2=0,3.22,4=6,72l

=>mHcl=0,6.36,5=21,9g

=>mdd=219g