PTPU

CuSO₄+ Ba(OH)₂\(\rightarrow\) BaSO₄\(\downarrow\)+ Cu(OH)₂\(\downarrow\)

1: 1: 1: 1

ta có: n_CuSO4= 0,2. 1= 0,2( mol)

n_Ba(OH)2= 0,3. 0,5= 0,15( mol)

ta có tỉ lệ: \(\dfrac{0,2}{1}\)> \(\dfrac{0,15}{1}\)

\(\Rightarrow\) CuSO₄ dư, Ba(OH)₂ hết

theo PTPU có: n_{CuSO4 pư}= n_BaSO4= n_Cu(OH)2= n_Ba(OH)2= 0,15( mol)

\(\Rightarrow\) m_{chất rắn sau pư}= m_BaSO4+ m_Cu(OH)2

= 0,15. 233+ 0,15. 98= 49,65( g)

ta có: n_{CuSO4 dư}= 0,2- 0,15= 0,05( mol)

\(\Rightarrow\) C_{M CuSO4}= \(\dfrac{0,05}{0,2+0,3}\)= 0,1M

Cu(OH)₂\(\xrightarrow[]{to}\) CuO+ H₂O

..0,15............0,15............ mol

\(\Rightarrow\) m_{chất rắn sau pư}= m_BaSO4+ m_CuO

= 0,15. 233+ 0,15. 80= 46,95( g)

đổi 200ml = 0,2l , 300ml = 0,3l

nCuSO4=0,2.1=0,2mol

nBa(OH)2=0,3.0,5=0,15mol

a)

pt : CuSO4 + Ba(OH)2 -----> Cu(OH)2 + BaSO4\(\downarrow\)

ncó: 0,2 0,15

npứ: 0,15 <---- 0,15 -------> 0,15 --------> 0,15

n dư: 0,05 0

b) chất CuSO4 dư , Ba(OH)2 hết

c) mBaSO4=0,15.233=34,95g

d)

Vdd sau pứ = Vdd CuSO4 + Vdd Ba(OH)2

= 0,2+0,3=0,5l

CM(CuSO4 dư)= 0,15/0,5=0,3M

CM(Cu(OH)2) = 0,15/0,5=0,3M

e)

pt: 2BaSO4 ---to---> 2BaO + 2SO2 +O2

n pứ : 0,15 ------------> 0,15

mBaO = 0,15. 153=22,95g

\(CuCl_2+2NaOH--->Cu\left(OH\right)_2\downarrow+2NaCl\left(1\right)\)

0,1_______0,3x______________0,1________0,2

\(Cu\left(OH\right)_2--to->CuO+CO_2\uparrow\left(2\right)\)

0,1_______________0,1

\(n_{CuCl_2}=0,2.0,5=0,1\left(mol\right)\)

\(n_{NaOH}=0,3x\left(mol\right)\)

b) =>\(0,3x=0,1.2=>x=0,67\left(M\right)\)

=> \(m=0,1.80=8\left(g\right)\)

c) => \(C_{M_{NaCl}}=\frac{0,2}{0,5}=0,4\left(M\right)\)

a)

$Fe_2(SO_4)_3 + 6KOH \to 2Fe(OH)_3 + 3K_2SO_4$

b)

$n_{Fe_2(SO_4)_3} = 0,3.1 = 0,3(mol)$
$n_{KOH} = \dfrac{16,8}{56} =0,3(mol)$

Ta thấy : 

$n_{KOH} : 3 < n_{Fe_2(SO_4)_3} : 1$ nên $Fe_2(SO_4)_3$ dư

$n_{Fe(OH)_3} = \dfrac{1}{3}n_{KOH} = 0,1(mol)$
$n_{Fe_2O_3} = \dfrac{1}{2}n_{Fe(OH)_3} = 0,05(mol)$

$m_{Fe_2O_3} = 0,05.160 = 8(gam)$

\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)

Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)

Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)

\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)

500ml=0.5l

nBaOH2 =0.5 x1=0.5 mol

MH2SO4=500.15%=75g

nH2SO4= xấp xỉ 0.8mol

H2SO4 dư tính theo BaOH2

pthh: Ba(OH)2 + H2SO4 => BaSO4+H2O

Theo pthh nBaSO4= nBa(OH)2=0.5mol

=>m kết tủa= 0.5x233=116.5g

theo pthh nH2SO4 phản ứng=nBaOH2= 0.5 mol

=> nH2SO4 Dư=0.8-0.5=0.3 mol

=>

m dư=0.3x98=29.4g

mH2SO4 đã dùng là m phản ứng? nếu thế thì m đã dung là 75-29.4=45.6

còn nếu m đã dùng là m chất tan thi là 75g như trên =))

\(a)CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\\n_{CaCl_2}=\dfrac{11}{111}=0,1\left(mol\right)\\ n_{AgCl}= 2n_{CaCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{AgCl}=0,2.143,5=28,7\left(g\right)\\ b)n_{AgNO_3}=2n_{CaCl_2}=0,2\left(mol\right)\\ n_{Ca\left(NO_3\right)_2}=n_{CaCl_2}=0,1\left(mol\right)\\ CM_{AgNO_3}=\dfrac{0,2}{0,035}=5,7M\\ CM_{Ca\left(NO_3\right)_2}=\dfrac{0,1}{15+35}=2M\)

nCuSO4 = 0.01 mol

Fe + CuSO4 --> FeSO4 + Cu

______0.01______0.01___0.01

X : Fe dư , Cu

Fe + 2HCl --> FeCl2 + H2

mCu = 0.01*64 = 0.64 (g)

FeSO4 + 2NaOH --> Fe(OH)2 + Na2SO4

0.01_______0.02

VddNaOH = 0.02/1 = 0.02 (l)

Tham khảo:

nCuSO4 = \(0,01.1=0,01\left(mol\right)\)

Fe + CuSO4 --> FeSO4 + Cu

........0,01..............0,01......0,01(mol)

Vì Fe,Cu dư

Fe + 2HCl --> FeCl2 + H2

mCu=0,64(g)

FeSO4 + 2NaOH --> Fe(OH)2 + Na2SO4

0.01...........0,02(mol)

V\(_{ddNaOH}=0,02.1=0,02\left(l\right)\)

#Walker

PTHH: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

Ta có: \(n_{Ba\left(OH\right)_2}=0,5\cdot1=0,5\left(mol\right)=n_{BaSO_4}=n_{H_2SO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,5\cdot98}{15\%}\approx326,7\left(g\right)\\m_{BaSO_4}=0,5\cdot233=116,5\left(g\right)\end{matrix}\right.\)

\(n_{Ba\left(OH\right)_2}=0,5\cdot1=0,5mol\)

a)\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

    0,5                 0,5            0,5        

b) \(m_{H_2SO_4}=0,5\cdot98=49\left(g\right)\)

   \(\Rightarrow m_{ddH_2SO_4}=\dfrac{49}{15}\cdot100=326,67\left(g\right)\)

c) \(m_{BaSO_4}=0,5\cdot233=116,5\left(g\right)\)