\(n_{H_2SO_4}=0,05\cdot3=0,15mol\)

a) \(H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\)

     0,15         0,15         0,15            0,3

    \(m_{ctBaCl_2}=0,15\cdot208=31,5\left(g\right)\)

    \(m_{BaCl_2thamgia}=\dfrac{31,5}{20\%}\cdot100\%=157,5\left(g\right)\)

b) \(m_{BaSO_4}=0,15\cdot233=34,95\left(g\right)\)

c) \(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

      0,15             0,15

    \(m_{ctCa\left(OH\right)_2}=0,15\cdot74=11,1\left(g\right)\)

    \(m_{ddCa\left(OH\right)_2}=\dfrac{11,1}{25\%}\cdot100\%=44,4\left(g\right)\)

    \(\Rightarrow V_{Ca\left(OH\right)_2}=\dfrac{44,4}{1,15}=38,6\left(ml\right)\)

\(n_{Zn}=\dfrac{4,55}{65}=0,07(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ a,n_{HCl}=0,14(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,14}{0,2}=0,7M\\ b,n_{H_2}=0,07(mol)\\ \Rightarrow V_{H_2}=0,07.22,4=1,568(l)\\ c,n_{ZnCl_2}=0,07(mol)\\ \Rightarrow m_{ZnCl_2}=0,07.136=9,52(g)\\ c,ZnCl_2+2AgNO_3\to 2AgCl\downarrow+Zn(NO_3)_2\)

\(m_{dd_{ZnCl_2}}=200.0,8+4,55-0,07.2=164,41(g)\\ n_{AgCl}=0,14(mol);n_{Zn(NO_3)_2}=0,07(mol)\\ \Rightarrow C\%_{Zn(NO_3)_2}=\dfrac{0,07.189}{164,41+200-0,14.143,5}.100\%=3,84%\)

a, PT: \(Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\)

Ta có: \(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\)

Theo PT: \(n_{SO_2}=n_{Na_2SO_3}=0,1\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(n_{NaCl}=n_{HCl}=2n_{Na_2SO_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{NaCl}=0,2.58,5=11,7\left(g\right)\)

c, \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{7,3}{10\%}=73\left(g\right)\)

d, Ta có: m _{dd sau pư} = 12,6 + 73 - 0,1.64 = 79,2 (g)

\(\Rightarrow C\%_{NaCl}=\dfrac{11,7}{79,2}.100\%\approx14,77\%\)

a) Fe + H2SO4 -----------> FeSO4 + H2

\(n_{Fe}=n_{H_2}=0,75\left(mol\right)\)

=> \(m_{Fe}=0,75.56=42\left(g\right)\)

b) \(CM_{H_2SO_4}=\dfrac{0,75}{0,25}=3M\)

c) \(m_{ddsaupu}=42+250.1,1-0,75.2=315,5\left(g\right)\)

=> \(C\%_{FeSO_4}=\dfrac{0,75.152}{315,5}.100=36,13\%\)

\(n_{SO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)

\(PTHH:2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

 Mol:         0,5           1,5                   0,25          0,75       1,5

a)m_Fe=0,5.56=28 (g)

b)\(C_{MddH_2SO_4}=\dfrac{1,5}{0,25}=6\left(mol/l\right)\)

c)\(m_{Fe_2\left(SO_4\right)_3}=0,25.400=100\left(g\right)\)

  \(m_{H_2O}=1,5.18=27\left(g\right)\)

\(C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{100.100}{100+27}=78,74\%\)

Pls ai giúp mình đi cần gấp lắm

2AgNO3+BaCl2=2AgCl+Ba(NO3)2(1)

theo (1) nAgCl=nAgNO3=16.8/170=0.099mol

mdd = 16.8+300*1.33-0.099*143.5=401.5935g

Vdd=300ml

Ddd muối=m/V=401.5935/300=1.338645g/ml

\(PTHH:CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ n_{CO_2}=\dfrac{26,88}{22,4}=1,2\left(mol\right)\\ \Rightarrow n_{Na_2CO_3}=1,2\left(mol\right)\\ \Rightarrow m_{muối}=m_{Na_2CO_3}=1,2\cdot106=127,2\left(g\right)\)