nAgNO3= (100.17%)/170=0,1(mol)

nHCl= (300.3,65%)/36,5=0,3(mol)

a) PTHH: AgNO3 + HCl -> AgCl + HNO3

Ta có: 0,1/1 < 0,3/1

=> AgNO3 hết, HCl dư, tính theo nAgNO3

Ta có: nAgCl= nHNO3= nHCl(p.ứ)= nAgNO3= 0,1(mol)

=>m(kt)=mAgCl= 143,5.0,1= 14,35(g)

b) mHCl(dư)= (0,3- 0,1).36,5=7,3(g)

mHNO3= 63.0,1= 6,3(g)

mddsau= mddAgNO3 + mddHCl - mAgCl= 100+300- 14,35= 385,65(g)

=>C%ddHCl(dư)= (7,3/385,65).100= 1,893%

C%ddHNO3= (6,3/385,65).100=1,634%

a)

$AgNO_3 + HCl \to AgCl + HNO_3$
Theo PTHH : 

$n_{AgCl} = n_{HCl} = n_{AgNO_3} = \dfrac{340.10\%}{170} =0,2(mol)$

$m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)$

b)

$m_{AgCl} = 0,2.143,5 = 28,7(gam)$

c)

$m_{dd\ sau\ pư} = 340 + 100 -28,7 = 411,3(gam)$

$n_{HNO_3} = n_{AgNO_3} = 0,2(mol)$
$\Rightarrow C\%_{HNO_3} = \dfrac{0,2.63}{411,3}.100\% = 3,06\%$

a)PTHH:     AgNO₃  + HCl → AgCl↓ + HNO₃ 

nHCl = 0,2.0,5 = 0,1 mol 

=> nAgCl = 0,1 mol  = nAgNO₃ = 0,1 mol  = nHCl phản ứng 

<=> mAgCl = 0,1.143,5 = 14,35 gam

mAgNO₃ = 0,1.170  = 17 gam

=> mdd AgNO₃  = \(\dfrac{17}{6,8\%}\)= 250 gam

b)    X +    2HCl --> XCl₂   + H₂ 

1,2 gam X tác dụng vừa đủ với 0,1 mol HCl

=> Số mol của 1,2 gam X = 0,05 mol

<=> M_x = \(\dfrac{1,2}{0,05}\)= 24  (g/mol)   =>   X là magie ( Mg )

n_CuCl2 = \(\dfrac{270.15\%}{100\%.135}\)= 0,3(mol)

CuCl₂ + 2KOH ➝ Cu(OH)₂↓ + 2KCl

  0,3  ➝    0,6    ➝     0,3        ➝  0,6         (mol)

a, m_Cu(OH)2 = 0,3.98= 29,4(g)

b,  m _{dd KOH} = \(\dfrac{0,6.56.100\%}{20\%}\)= 168(g)

c, m_KCl = 0,6.74,5 = 44,7(g)(*)

   Áp dụng định luật bảo toàn khối lượng:

=> m_{dd KCl}= m_CuCl2 + m _{dd KOH} - m_Cu(OH)2

_⇔ m_{dd KCl} = 0,3.135+ 168 - 29,4 = 179,1(g)(**)

Từ (*) và (**) ⇒  C%_KCl = \(\dfrac{44,7}{179,1}\).100%\(\approx\) 24,96%

n_CuCl2 = \(\dfrac{270.15\%}{100\%.135}\)= 0,3(mol)

CuCl₂ + 2KOH ➝ Cu(OH)₂↓ + 2KCl

  0,3  ➝    0,6    ➝     0,3        ➝  0,6         (mol)

a, m_Cu(OH)2 = 0,3.98= 29,4(g)

b, m_KOH = 0,6.56 = 33,6(g)

c, m_KCl = 0,6.74,5 = 44,7(g)(*)

   m _{dd KOH} = \(\dfrac{33,6.100\%}{20\%}\)= 168(g)

Áp dụng định luật bảo toàn khối lượng:

=> m_{dd KCl}= m_CuCl2 + m _{dd KOH} - m_Cu(OH)2

_⇔ m_{dd KCl} = 0,3.135+ 168 - 29,4 = 179,1(g)(**)

Từ (*) và (**) ⇒  C%_KCl = \(\dfrac{44,7}{179,1}\).100%\(\approx\) 24,96%

PTHH: \(CuCl_2+2KOH\rightarrow2KCl+Cu\left(OH\right)_2\downarrow\)

a) Ta có: \(n_{CuCl_2}=\dfrac{270\cdot15\%}{135}=0,3\left(mol\right)=n_{Cu\left(OH\right)_2}\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,3\cdot98=29,4\left(g\right)\)

b) Theo PTHH: \(n_{KOH}=2n_{CuCl_2}=0,6mol\) \(\Rightarrow m_{ddKOH}=\dfrac{0,6\cdot57}{20\%}=171\left(g\right)\)

c) Theo PTHH: \(n_{KCl}=n_{KOH}=0,6mol\) \(\Rightarrow m_{KCl}=0,6\cdot74,5=44,7\left(g\right)\)

Mặt khác: \(m_{dd}=m_{ddCuCl_2}+m_{ddKOH}-m_{Cu\left(OH\right)_2}=411,6\left(g\right)\)

\(\Rightarrow C\%_{KCl}=\dfrac{44,7}{411,6}\cdot100\%\approx10,86\%\)

\(n_{CuSO_4}=\dfrac{200.16\%}{160}=0,2\left(mol\right)\)

PTHH :

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

0,2              0,4                  0,2               0,2 

\(m_{NaOH}=0,4.40=16\left(g\right)\)

\(m_{ddNaOH}=\dfrac{16.100}{10}=160\left(g\right)\)

\(c,m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)

\(m_{ddNa_2SO_4}=200+160-\left(0,2.98\right)=340,4\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{28,4}{240,4}.100\%\approx8,34\%\)

\(d,PTHH:\)

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

0,2               0,2 

\(m_{CuO}=0,2.80=16\left(g\right)\)

a, \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

b, \(m_{CuSO_4}=200.16\%=32\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)

\(n_{NaOH}=2n_{CuSO_4}=0,4\left(mol\right)\Rightarrow m_{ddNaOH}=\dfrac{0,4.40}{10\%}=160\left(g\right)\)

c, \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,2\left(mol\right)\)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,2.142}{200+160-0,2.98}.100\%\approx8,34\%\)

d, \(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

\(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

PTHH: \(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\downarrow\)

Ta có: \(n_{NaCl}=0,2\cdot0,5=0,1\left(mol\right)=n_{AgNO_3}=n_{AgCl}\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{AgNO_3}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\m_{AgCl}=0,1\cdot143,5=14,35\left(g\right)\end{matrix}\right.\)