Trả lời:

\(n_{FeO}=\dfrac{10,8}{72}=0,15\left(mol\right)\)

\(m_{HCl}=\dfrac{150.18,25}{100}=27,375\left(g\right)=>n_{HCl}=\dfrac{27,375}{36,5}=0,75\left(mol\right)\)

PTHH :

\(FeO+2HCl-->FeCl_2+H_2O\)

\(\dfrac{0,15}{1}< \dfrac{0,75}{2}\)

\(=>FeO\) hết , \(HCl\) dư , tính theo FeO

a,

\(n_{FeCl_2}=n_{FeO}=0,15\left(mol\right)=>m_{FeCl_2}=0,15.127=19,05\left(g\right)\)

b, Sau phản ứng HCl dư nên :

\(m_{HCl_{dư}}=m_{HCl_{bđ}}-m_{HCl_{pư}}=36,5.0,75-36,5.0,3=16,425\left(g\right)\)

c, H\(_2\)SO\(_4\) phản ứng với FeO

\(FeO+H_2SO_4-->FeSO_4+H_2O\)

\(n_{H_2SO_4}=n_{FeO}=0,15mol=>m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)

\(=>m_{ddH_2SO_4}=\dfrac{14,7.100}{20}=73,5\left(g\right)\)

\(n_{Ca}=\dfrac{8}{40}=0,2mol\\ 2Ca+O_2\xrightarrow[]{t^0}2CaO\\ n_{CaO}=n_{Ca}=0,2mol\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5mol\\ CaO+2HCl\rightarrow CaCl_2+H_2O\\ \Rightarrow\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ n_{CaCl_2}=n_{CaO}=0,2mol\\ n_{HCl}=2n_{CaO}=0,4mol\\ m_{CaCl_2}=0,2.111=22,2g\\ m_{HCl.dư}=\left(0,5-0,4\right).36,5=3,65g\)

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(n_{HCl}=\dfrac{182.5\cdot10}{100\cdot36.5}=0.5\left(mol\right)\)

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

\(0.2......0.4..........0.2........0.2\)

\(n_{HCl\left(dư\right)}=0.5-0.4=0.1\left(mol\right)\)

\(m_{HCl\left(dư\right)}=0.1\cdot36.5=3.65\left(g\right)\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=13+182.5-0.2\cdot2=195.1\left(g\right)\)

\(C\%_{HCl\left(dư\right)}=\dfrac{3.65}{195.1}\cdot100\%=1.87\%\)

\(C\%_{ZnCl_2}=\dfrac{0.2\cdot136}{195.1}\cdot100\%=13.94\%\)

\(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\ m_{H_2SO_4}=\dfrac{22,05.20}{100}=4,41\left(g\right)\\ n_{H_2SO_4}=\dfrac{4,41}{98}=0,045\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\) 
\(LTL:\dfrac{0,03}{1}< \dfrac{0,045}{1}\) 
=> H₂SO₄ dư 
\(n_{H_2SO_4\left(p\text{ư}\right)}=n_{ZnSO_4}=n_{H_2}=n_{Zn}=0,03\left(mol\right)\\ m_{H_2SO_4\left(d\right)}=\left(0,045-0,03\right).98=1,47\left(g\right)\\ m_{\text{dd}}=1,95+22,05-\left(0,03.2\right)=23,94\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,03.136}{23,94}.100\%=17\%\)

\(a,n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\\ n_{H_2SO_4}=\dfrac{22,05}{98}=0,225\left(mol\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

bđ       0,03   0,225

pư       0,03    0,03 

spư     0         0,195       0,03        0,03

\(b,m_{H_2SO_4\left(dư\right)}=0,195.98=19,11\left(g\right)\\ c,m_{dd}=1,95+22,05-0,03.2=23,94\left(g\right)\\ C\%_{ZnSO_4}=\dfrac{0,03.161}{23,94}.100\%=20,18\%\)

a)b)c)d) mBaCl2=150.16,64%=24,96g

=>nBaCl2=0,12 mol

mH2SO4=100.14,7%=14,7g=>nH2SO4=0,15mol

     BaCl2       + H2SO4 =>BaSO4    +2HCl

Bđ: 0,12 mol;    0,15 mol

Pứ: 0,12 mol=>0,12 mol=>0,12 mol=>0,24 mol

Dư:                   0,03 mol

Dd ban đầu chứa BaCl2 0,12 mol và H2SO4 0,15 mol

Dd A sau phản ứng chứa HCl 0,24 mol và H2SO4 dư 0,03 mol

mHCl=0,24.36,5=8,76g

mH2SO4=0,03.98=2,94g

Kết tủa B là BaSO4 0,12 mol=>mBaSO4=0,12.233=27,96g

mddA=mddBaCl2+mddH2SO4-mBaSO4

=150+100-27,96=222,04g

C%dd HCl=8,76/222,04.100%=3,945%

C% dd H2SO4=2,94/222,04.100%=1,324%

e) HCl     +NaOH =>NaCl +H2O

0,24 mol=>0,24 mol

H2SO4 +2NaOH =>Na2SO4 + 2H2O

0,03 mol=>0,06 mol

TÔNG nNaOH=0,3 mol

=>V dd NaOH=0,3/2=0,15 lit

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ a,Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=2.0,2=0,4\left(mol\right)\\ b,m_{ddHCl}=\dfrac{0,4.36,5.100}{20}=73\left(g\right)\\ c,n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\ m_{ddsau}=11,2+73-0,2.2=83,8\left(g\right)\\ C\%_{ddFeCl_2}=\dfrac{0,2.127}{83,8}.100\approx30,31\%\)