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

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)

\(n_{CuCl_2}=\dfrac{60,75}{135}=0,45mol\\ a)CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
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\(\left(Cu\left(OH\right)_2\right)\)

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

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

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

\(SO_3+H_2O\rightarrow H_2SO_4\)

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

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

\(b.\)

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

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

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

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

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

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

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, \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

b, \(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

Theo PT: \(n_{CuCl_2}=n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,125\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)

c, \(C_{M_{CuCl_2}}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

a) PTHH : \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

                   0,125<-----0,25---------------------->0,125

b) \(m_{Kt}=m_{Cu\left(OH\right)2}=0,125.98=12,25\left(g\right)\)

c) \(C_{MCuCl2}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

PTHH:

\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)

0,125       0,25                0,125         0,25

\(m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)

\(C_{M\left(CuCl_2\right)}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,15<-0,3<----0,15<---0,15

\(\left\{{}\begin{matrix}\%Fe=\dfrac{0,15.56}{12}.100\%=70\%\%\\\%Cu=100\%-70\%=30\%\end{matrix}\right.\)

c) m_HCl = 0,3.36,5 = 10,95 (g)

=> \(m_{dd}=\dfrac{10,95.100}{10}=109,5\left(g\right)\)

d) mdd  = 12 + 109,5 - 0,15.2 = 121,2 (g)

 \(C\%\left(FeCl_2\right)=\dfrac{0,15.127}{121,2}.100\%=15,718\%\)

a: \(Cu+2HCl->CuCl_2+H_2\)

\(Fe+2HCl->FeCl_2+H_2\)