a) BaCl2 + H2SO4 --> BaSO4 + 2HCl (1)

nH2SO4=0,0286(mol)

nBaCl2=0,1(mol)

lập tỉ lệ :

\(\dfrac{0,1}{1}>\dfrac{0,0286}{1}\)

=> BaC2 dư ,H2SO4 hết => bài toán tính theo H2SO4

theo (1) : nBaCl2(pư)=nBaSO4=nH2SO4=0,0286(mol)

=>mBaSO4=6,6638(g)

b) mdd sau pư = 14+400-6,6638=407,3362(g)

theo (1) : nHCl=2nH2SO4=0,0572(mol)

=>C%dd BaCl2(dư)=(0,1-0,0286).208/407,3362.100=3,65(%)

C%dd HCl=0,5125(%)

Cho 28 gam chất gì em nhỉ? Sắt hả?

a) PTHH: H₂SO₄ + BaCl₂ ➜ BaSO₄↓ + 2HCl

b) \(m_{H_2SO_4}=49\times20\%=9,8\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

\(m_{BaCl_2}=200\times5,2\%=10,4\left(g\right)\)

\(\Rightarrow n_{BaCl_2}=\dfrac{10,4}{208}=0,05\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{BaCl_2}\)

Theo bài: \(n_{H_2SO_4}=2n_{BaCl_2}\)

Vì \(2>1\) ⇒ dd H₂SO₄ dư, dd BaCl₂ hết

Kết tủa A gồm: BaSO₄

Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,05\left(mol\right)\)

\(\Rightarrow m_{BaSO_4}=0,05\times233=11,65\left(g\right)\)

c) Dung dịch B gồm: HCl, H₂SO₄ dư

Theo PT: \(n_{H_2SO_4}pư=n_{BaCl_2}=0,05\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}dư=0,1-0,05=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}dư=0,05\times98=4,9\left(g\right)\)

Theo PT: \(n_{HCl}=2n_{BaCl_2}=2\times0,05=0,1\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,1\times36,5=3,65\left(g\right)\)

\(\Sigma m_{ddB}=49+200=249\left(g\right)\)

\(C\%_{ddH_2SO_4}dư=\dfrac{4,9}{249}\times100\%\approx1,97\%\)

\(C\%_{ddHCl}=\dfrac{3,65}{249}\times100\%\approx1,47\%\)

\(a)n_{H_2SO_4}=\dfrac{58,8.20}{100.98}=0,12mol\\ n_{BaCl_2}=\dfrac{200.5,2}{100.208}=0,05mol\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ \Rightarrow\dfrac{0,12}{1}>\dfrac{0,05}{2}\Rightarrow H_2SO_4.dư\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

0,05         0,05             0,05           0,1

\(m_{BaSO_4}=0,05.233=11,65g\\ b)m_{dd}=58,8+200-11,65=247,15g\\ C_{\%HCl}=\dfrac{0,1.36,5}{247,15}\cdot100=1,48\%\\ C_{\%H_2SO_4,dư}=\dfrac{\left(0,12-0,05\right).98}{247,15}\cdot100=2,78\%\)

a/

\(n_{Na_2O}=\dfrac{9,3}{62}=0,15\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,15 0,3 (mol)

\(m_{NaOH}=0,3.40=12\left(g\right)\)

\(m_A=90,7+9,3=100\left(g\right)\)

\(C\%_{NaOH}=\dfrac{12}{100}.100\%=12\%\)

b/

m\(_{FeSO_4}=\dfrac{16.200}{100}=32\left(g\right)\)

\(\rightarrow m_{FeSO_4}=\dfrac{32}{152}=\dfrac{4}{19}\left(mol\right)\)

\(2NaOH+FeSO_4\rightarrow Na_2SO_4+Fe\left(OH\right)_2\downarrow\)

bđ: 0,3 \(\dfrac{4}{19}\) 0 0 (mol)

pư: 0,3 0,15 0,15 0,15 (mol)

dư: 0 \(\dfrac{23}{380}\) (mol)

\(m_{Fe\left(OH\right)_2}=0,15.90=13,5\left(g\right)\)

\(m_C=100+200-13,5=286,5\left(g\right)\)

\(m_{Na_2SO_4}=0,15.142=21,3\left(g\right)\)

\(\rightarrow C\%_{Na_2SO_4}=\dfrac{21,3}{286,5}.100\%\approx7,4\%\)

\(m_{FeSO_4\left(dư\right)}=\dfrac{23}{380}.152=9,2\left(g\right)\)

\(\rightarrow C\%_{FeSO_4\left(dư\right)}=\dfrac{9,2}{286,5}.100\%\approx3,2\%\)

\(n_{BaCl_2}=\frac{400.5,2\%}{208}=0,1\left(mol\right);n_{H_2SO_4}=\frac{100.1,14.19,6\%}{98}=0,228\left(mol\right)\)

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

Theo đề: 0,1.........0,228.....................................(mol)

Lập tỉ lệ: \(\frac{0,1}{1}< \frac{0,228}{1}\)=> Sau phản ứng H2SO4 dư

Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

dd sau khi lọc bỏ kết tủa: H2SO4 dư, HCl

\(m_{ddsaup.ứ}=400+114-23,3=490,7\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4\left(dư\right)}=\frac{\left(0,228-0,1\right).98}{490,7}.100=2,56\%\)

\(C\%_{HCl}=\frac{0,1.2.36,5}{490,7}.100=1,49\%\)