a, \(H_2SO_4+BaCl_2\rightarrow BaSO_{4\downarrow}+2HCl\)

b, Ta có: \(m_{H_2SO_4}=114.20\%=22,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{22,8}{96}=0,2375\left(mol\right)\)

\(m_{BaCl_2}=400.5,2\%=20,8\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2375}{1}>\dfrac{0,1}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=n_{BaCl_2}=0,1\left(mol\right)\\n_{HCl}=2n_{BaCl_2}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,2375-0,1=0,1375\left(mol\right)\)

Ta có: m _{dd sau pư} = 114 + 400 - 23,3 = 490,7 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1375.98}{490,7}.100\%\approx2,75\%\\C\%_{HCl}=\dfrac{0,2.36,5}{490,7}.100\%\approx1,49\%\end{matrix}\right.\)

K/l H2SO4:

......mH2SO4=114.20%=22,8(g)

mBaCl2=400.5,2%=20,8(g)

PTHH: H2SO4+BaCl2--->2HCl+ BaSO4

Có: 22,8/(2+32+16.4)>20,8/(137+35,5.2)=> H2SO4 dư \(13g\)

PTHH: H2SO4+BaCl2--->2HCl+ BaSO4

.................9,8......20,8..........7,3.........23,3(g)

Có: m(dd sau p/ứ)=114+400=514(g)

\(C\%_{H_2SO_4}=\frac{13}{514}.100=\frac{650}{257}\left(\%\right)\)

\(C\%_{HCl}=\frac{7,3}{514}.100=\frac{365}{257}\left(\%\right)\)

\(C\%_{BaSO_4}=\frac{23,3}{514}.100=\frac{1165}{257}\left(\%\right)\)

#Walker

Theo đề bài ta có :

\(\left\{{}\begin{matrix}nH2SO4=\dfrac{\left(100.1,14\right).20}{98.100}\approx0,23\left(mol\right)\\nBaCl2=\dfrac{400.5,2}{100.208}=0,1\left(mol\right)\end{matrix}\right.\)

Ta có PTHH :

\(BaCl2+H2SO4->B\text{aS}O4\downarrow+2HCl\)

0,1mol.......0,1mol.............0,1mol.......0,2mol

Theo PTHH ta có : \(nBaCl2=\dfrac{0,1}{1}mol< nH2SO4=\dfrac{0,23}{1}mol\)

=> nH2SO4 dư ( tính theo nBaCl2)

a) Ta có : mBaSO4 = 0,1.233=23,3(g)

b) Ta có : \(\left\{{}\begin{matrix}C\%H2SO4\left(d\text{ư}\right)=\dfrac{\left(0,23-0,1\right).98}{100.1,14+400-23,3}.100\%\approx2,596\%\\C\%HCl=\dfrac{0,2.36,5}{100.1,14+400-23,3}.100\%\approx1,49\%\end{matrix}\right.\)

Vậy.......

n_BaCl2= 0,1 (mol)

n_H2SO4 = 0,2327 (mol)

BaCl₂ + H₂SO₄ \(\rightarrow\) BaSO₄ \(\downarrow\) + 2HCl

bđ 0,1 0,2327 }

pư 0,1 \(\rightarrow\) 0,1 \(\rightarrow\) 0,1 \(\rightarrow\) 0,2 } (mol)

spư 0 0,1327 0,1 0,2 }

m_BaSO4 = 0,1 . 233 = 23,3 (g)

mdd(sau pư)= 400 + 1,14 . 100 - 23,3 =490,7 (g)

C%_(H2SO4)=\(\frac{0,1327.98}{490,7}\) . 100% = 2,65%

C% (HCl) =\(\frac{0,2.36,5}{490,7}\) . 100% = 1,49%

a.) \(PT:\) \(BaCl_2+H_2SO_4->BaSO_4+2HCl\)

\(m_{ddH_2SO_4}=100.1,14=114g\)

\(m_{ctH_2SO_4}=\dfrac{114.20}{100}=22,8g\)

\(nH_2SO_4=\dfrac{22,8}{98}\approx0,233mol\)

\(mct_{BaCl_2}=\dfrac{400.5,2}{100}=20,8g\)

\(nBaCl_2=\dfrac{20,8}{208}=0,1mol\)

PT: \(BaCl_2+H_2SO_4->BaSO_4+2HCl\)

0,1 0,1 0,2 (mol)

Theo số mol của H₂SO₄ => nBaCl sẽ là 0,233mol

Theo số mol của BaCl => nH₂SO₄ sẽ là 0,1 mol

=> H2SO4 dư

Sau PƯ có chất kết tủa trắng BaSO4

=> \(mBaCl=0,1.233=23,3g\)

b.)

Dung dịch sau PƯ có chứa: HCl: 0,2mol

H2SO4 dư: 0,234 - 0,1 = 0,133mol

\(mH2SO4\) dư \(=96.0,133=13,034g\)

\(mHCl=0,2.36,5=7,3g\)

\(mdd\) sau PƯ \(=114+400-23,3=490,7g\)

\(\Rightarrow C\%_{HCl}=\dfrac{7,3}{490,7}.100\approx1,49\%\)

\(\Rightarrow C\%_{H2SO4dư}=\dfrac{13.034}{490,7}.100\approx2,66\%\)

mBaCl2 = 400.5,2% = 20,8g 
mH2SO4 = 100.1,14.20% = 22,8g 

BaCl2 + H2SO4 = BaSO4 + 2HCl 
208...... 98........... 233........ 2.36,5g 
20,8..... 22,8......... 

20,8/208 < 22,8/98 => BaCl2 phản ứng hết và H2SO4 còn dư 
=> mBaSO4 = 20,8.233/208 = 23,3g 
=> mHCl = 20,8.2.36,5/208 = 7,3g 
=> mH2SO4 (dư) = 22,8 - 20,8.98/208 = 13g 

mdd (trước ph.ư) = 400 + 100.1,14 = 514g 
mdd (sau ph.ư) = 514 - 23,3 = 490,7g 

=> C%HCl = 7,3/490,7 = 1,49% 
=> C%H2SO4 (dư) = 13/490,7 = 2,65% 

nbacl2= 400*5,2/(100*208)=0,1 mol

mdd h2so4= 100*1,14=114g

nh2so4=114*20/(100*98)=57/245 mol

bacl2+ h2so4-> baso4+ 2hcl

thấy 0,1< 57/245

=> sau pư bacl2 hết, h2so4 còn dư

nh2so4 dư=57/245-0,1=13/98

n baso4= nbacl2= 0,1 mol

=> mbaso4= 0,1*233=23,3g

mdd sau pư=400+114-23,3=490,7g

C% h2so4 dư=13/98*98*100/490,7=2,65%

C% hcl=0,1*2*36,5*100/490,7=1,49%

\(m_{BaCl_2}=\dfrac{200.5,2}{100}=10,4\left(g\right)\\ \rightarrow n_{BaCl_2}=\dfrac{10,4}{208}=0,05\left(mol\right)\)

\(m_{H_2SO_4}=\dfrac{58,8.20}{100}=11,76\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{11,76}{98}=0,12\left(mol\right)\)

\(PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

- Ta có: 0,05/1 < 0,12/1

=> BaCl2 hết, H2SO4 dư.

=> Các chất trong dd sau phản ứng là H2SO4 (dư) và HCl.

\(n_{BaSO_4}=n_{BaCl_2}=0,05\left(mol\right)\\ \rightarrow m_{BaSO_4}=0,05.233=11,65\left(g\right)\)

Ta có: \(m_{ddsau}=200+58,8-11,65=247,15\left(g\right)\)

\(m_{H_2SO_4\left(dư\right)}=11,76-\left(0,12-0,05\right).98=4,9\left(g\right)\)

\(n_{HCl}=2.0,05=0,1\left(mol\right)\\ \rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\)

=> \(C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{4,9}{247,15}.100\approx1,983\%\)

\(C\%ddHCl=\dfrac{3,65}{217,15}.100\approx1,477\%\)