\(CaCl_2+Na_2SO_4\rightarrow CaSO_4+2NaCl\)

\(n_{CaCl_2}=\dfrac{22,2}{111}=0,2\left(mol\right)\)

a) Đề sai

b) \(m_{ddNa_2SO_4}=200.1,55=310\left(g\right)\)

CaSO4 ít tan, coi như không tan, là kết tủa

\(m_{CaSO_4}=0,2.136=27,2\left(g\right)\)

Muối sau phản ứng là NaCl

\(m_{ddsaupu}=22,2+310-27,2=305\left(g\right)\)

\(C\%_{NaCl}=\dfrac{0,2.2.58,5}{305}.100=7,67\%\)

\(n_{CaCl_2}=\dfrac{22.2}{111}=0.2\left(mol\right)\)

\(m_{dd_{Na_2SO_3}}=200\cdot1.55=310\left(g\right)\)

\(CaCl_2+Na_2SO_3\rightarrow CaSO_3+2NaCl\)

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

\(m_{CaSO_3}=0.2\cdot120=24\left(g\right)\)

\(m_{NaCl}=0.4\cdot58.5=23.4\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=22.2+310-24=308.2\left(g\right)\)

\(C\%_{NaCl}=\dfrac{23.4}{308.2}\cdot100\%=7.59\%\)

\(n_{FeO}=\dfrac{14,4}{72}=0,2\left(mol\right)\)

\(m_{ct}=\dfrac{10,95.200}{100}=21,9\left(g\right)\)

\(n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Pt : \(FeO+2HCl\rightarrow FeCl_2+H_2O|\)

         1           2             1            1

        0,2       0,6            0,2

a) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,6}{2}\)

             ⇒ FeO phản ứng hết , HCl dư

             ⇒ Tính toán dựa vào số mol của FeO

\(n_{HCl\left(dư\right)}=0,6-\left(0,2.2\right)=0,2\left(mol\right)\)

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

b) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{FeCl2}=0,2.127=25,4\left(g\right)\)

\(m_{ddspu}=14,4+200=214,4\left(g\right)\)

\(C_{FeCl2}=\dfrac{25,4.100}{214,4}=11,85\)⁰/₀

\(C_{HCl\left(dư\right)}=\dfrac{7,3.100}{214,4}=3,4\)⁰/₀

 Chúc bạn học tốt

a,\(n_{FeO}=\dfrac{14,4}{72}=0,2\left(mol\right)\)

 \(m_{HCl}=10,95\%.200=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

PTHH: FeO + 2HCl → FeCl₂ + H₂O

Mol:      0,2      0,4         0,2

Ta có: \(\dfrac{0,2}{1}< \dfrac{0,6}{2}\) ⇒ FeO pứ hết ,HCl dư

m_{HCl dư} = (0,6-0,4).36,5 = 7,3 (g)

b, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

c, m_{dd sau pứ} = 14,4+200 = 214,4 (g)

\(C\%_{ddFeCl_2}=\dfrac{25,4.100\%}{214,4}=11,847\%\)

\(C\%_{ddHCldư}=\dfrac{7,3.100\%}{214,4}=3,4\%\)

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

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

\(0.1......................0.1..........0.1\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{dd_{H_2SO_4}}=150\cdot1.08=162\left(g\right)\)

\(m_{dd}=6.5+162-0.1\cdot2=168.3\left(g\right)\)

\(C\%_{ZnSO_4}=\dfrac{0.1\cdot161}{168.3}\cdot100\%=9.56\%\)

Đổi: 672 ml = 0,672 l

CaSO₄ + H₂SO₄ → X

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + CO₂ + H₂O (1)

\(n_{CO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

a) Theo PT1: \(n_{H_2SO_4}=n_{CO_2}=0,03\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,03}{0,2}=0,15\left(M\right)\)

b) Theo PT: \(n_{Na_2SO_4}=n_{CO_2}=0,03\left(mol\right)\)

\(\Rightarrow m_{Na_2SO_4}=0,03\times142=4,26\left(g\right)\)

c) Theo PT1: \(n_{Na_2CO_3}=n_{CO_2}=0,03\left(mol\right)\)

\(\Rightarrow m_{Na_2CO_3}=0,03\times106=3,18\left(g\right)\)

\(\Rightarrow m_{CaSO_4}=5-3,18=1,82\left(g\right)\)

\(\Rightarrow\%m_{Na_2CO_3}=\dfrac{3,18}{5}\times100\%=63,6\%\)

\(\Rightarrow\%m_{CaSO_4}=100\%-63,6\%=36,4\%\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{Fe}=n_{H_2SO_4}=n_{Fe(OH)_2}=0,2(mol)\\ a,m_{Fe}=0,2.56=11,2(g)\\ b,C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1M\)

\(c,Ba(OH)_2+FeSO_4\to BaSO_4\downarrow+Fe(OH)_2\downarrow\\ n_{Ba(OH)_2}=\dfrac{250.17,1}{100.171}=0,25(mol)\\ LTL:\dfrac{0,2}{1}<\dfrac{0,25}{1}\Rightarrow Ba(OH)_2\text{ dư}\\ \Rightarrow n_{BaSO_4}=0,2(mol)\\ \Rightarrow m_{BaSO_4}=233.0,2=46,6(g)\)

PTHH: Zn + 2HCL ---> ZnCl2 + H2

a)Theo bài ta có:

nZn=mZn/MZn=13/65=0,2 mol

=> nHCl=2 nZn=2.0,2=0,4 mol

=>mHCl=n_HCl.M_HCl =0,4 . 36,5=14,6(g)

mdd HCl=D.Vdd=1,14 .200=228(g)

=> C% ddHCl=\(\dfrac{mHCl.100\%}{mddHCl}=\dfrac{14,6.100\%}{228}=6,4\%\)(xấp xỉ)

b) nZnCl2=nZn=0,2 mol

=> mZnCl2= nZnCl2.M_ZnCl2=0,2 .136=27,7(g)