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\(n_{MgCl_2}\)=\(0,1.2=0,2(mol)\)
\(n_{Ba(OH)_2}\)=\(0,15.1,5=0,225(mol) \)
\({MgCl_2}+{Ba(OH)_2}-->{Mg(OH)_2}+{BaCl_2}\)
Dung dịch A chứa 0,225-0,2=0,025 mol \({Ba(OH)_2}\) dư; 0,2 mol \({BaCl_2}\)
Kết tủa B là 0,2 mol \({Mg(OH)_2}\)
\({Mg(OH)_2}-->MgO+{H_2O}\)
⇒\(n_{MgO}\)=\(n_{Mg(OH)_2}=0,2 mol\)
⇒\(m_{MgO}=0,2.40=8(g)\)
Coi thể tích dung dịch không đổi sau khi trộn
\(V_{dd}=100+150=250ml=0,25l\)
⇒\(C_M{Ba(OH)_2}\)=\(\dfrac{0,025}{0,25}=0,1M\)
\(C_M{BaCl_2}=\dfrac{0,2}{0,25}=0,8M\)
mdd(sau phản ứng)=250.1,12=280(g)
C%\({Ba(OH)_2}=\dfrac{0,025.171}{280}.100=1,5%\)%
C%\({BaCl_2}=\dfrac{0,2.208}{280}.100=14,85%\)%
nMgCL2=0.2(mol)
nKOH=0.3(mol)
MgCL2+2KOH->Mg(OH)2+2KCl
0.2 0.3
->MgCl dư
nMg(OH)2=0.15(mol)CM=0.6(M)
nKCl=0.3(mol)CM=1.2(M)
nMgCl dư=0.2-0.3:2=0.05(mol)CM=0.2(M)
a.CuCl2 + 2NaOH -> Cu(OH)2 + 2NaCl
0.15 0.3 0.15 0.3
Cu(OH)2 -> CuO + H2O
0.15 0.15
nNaOH = 0.3 mol
\(CM_{CuCl2}=\dfrac{0.15}{2}=0.075M\)
b.Vdd sau phản ứng = 0.2 + 0.15 = 0.35l
\(CM_{NaCl}=\dfrac{0.3}{0.35}=0.86M\)
c.mCuO = \(0.15\times80=12g\)
nFe2(SO4)3=0,15(mol);
nBa(OH)2=0,3(mol)
Fe2(SO4)3+3Ba(OH)2--->3BaSO4+ 2Fe(OH)3
Xét 0,15/1>0,3/3 => Fe2(SO4)3dư , tính theo Ba(OH)2
theo pt nBa(OH)2=nBaSO4=0,3(mol)
nFe(OH)3=2/3nBa(OH)2=0,2
=> mkết tủa = 0,3.233+0,2.107=91,3(g)
b, dung dịch là Fe2(SO4)3
nFe2(SO4)3(pứ)=1/3nBa(OH)2=0,1(mol)
=> nFe2(SO4)3 dư = 0,15-0,1=0,05(mol)
Vdd=100+150=250(ml)=0,25(l)
=> CMFe2(SO4)3 = 0,2(M)
Bài 1:
a) CuSO4 + 2NaOH → Na2SO4 + Cu(OH)2↓
\(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)
Theo PT: \(n_{CuSO_4}=\dfrac{1}{2}n_{NaOH}\)
Theo bài: \(n_{CuSO_4}=\dfrac{1}{3}n_{NaOH}\)
Vì \(\dfrac{1}{3}< \dfrac{1}{2}\) ⇒ NaOH dư
b) Theo PT: \(n_{Cu\left(OH\right)_2}=m_{CuSO_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,1\times98=9,8\left(g\right)\)
c) \(\Sigma V_{dd}saupư=40+60=100\left(ml\right)=0,1\left(l\right)\)
Theo PT: \(n_{NaOH}pư=2n_{CuSO_4}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow n_{NaOH}dư=0,3-0,2=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}dư=\dfrac{0,1}{0,1}=1\left(M\right)\)
Theo PT: \(n_{Na_2SO_4}=n_{CuSO_4}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
Bài 2:
ZnCl2 + 2NaOH → 2NaCl + Zn(OH)2↓ (1)
\(n_{ZnCl_2}=0,3\times1,5=0,45\left(mol\right)\)
\(n_{NaOH}=0,1\times1=0,1\left(mol\right)\)
Theo PT1: \(n_{ZnCl_2}=\dfrac{1}{2}n_{NaOH}\)
Theo bài: \(n_{ZnCl_2}=\dfrac{9}{2}n_{NaOH}\)
Vì \(\dfrac{9}{2}>\dfrac{1}{2}\) ⇒ ZnCl2 dư
a) \(\Sigma V_{dd}saupư=300+100=400\left(ml\right)=0,4\left(l\right)\)
Theo PT1: \(n_{ZnCl_2}pư=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,1=0,05\left(mol\right)\)
\(\Rightarrow n_{ZnCl_2}dư=0,45-0,05=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{ZnCl_2}}dư=\dfrac{0,4}{0,4}=1\left(M\right)\)
Theo PT1: \(n_{NaCl}=n_{NaOH}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)
b) Zn(OH)2 \(\underrightarrow{to}\) ZnO + H2O (2)
Theo pT1: \(n_{Zn\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,1=0,05\left(mol\right)\)
Theo pT2: \(n_{ZnO}=n_{Zn\left(OH\right)_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{ZnO}=0,05\times81=4,05\left(g\right)\)
c) NaOH + HCl → NaCl + H2O (3)
Theo PT: \(n_{HCl}=n_{NaOH}=0,1\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,1\times36,5=3,65\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{3,65}{25\%}=14,6\left(g\right)\)
