PTHH: \(2AgNO_3+CaCl_2\rightarrow Ca\left(NO_3\right)_2+2AgCl\downarrow\)

Ta có: \(n_{AgNO_3}=\dfrac{1,7}{170}=0,01\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{AgCl}=0,01\left(mol\right)\\n_{CaCl_2}=n_{Ca\left(NO_3\right)_2}=0,005\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,005\cdot111=0,555\left(g\right)\\m_{AgCl}=0,01\cdot143,5=1,435\left(g\right)\\C_{M_{Ca\left(NO_3\right)_2}}=\dfrac{0,005}{0,07+0,03}=0,05\left(M\right)\end{matrix}\right.\)

ta có: m_CaCl2= 7,4. 45%= 3,33( g)

\(\Rightarrow\) n_CaCl2= \(\dfrac{3,33}{111}\)= 0,03( mol)

PTPU

CaCl₂+ 2AgNO₃\(\rightarrow\) Ca(NO₃)₂+ 2AgCl\(\downarrow\)

..0,03........0,06.............0,03............0,06.......... mol

\(\Rightarrow\) m_AgNO3= 0,06. 170= 10,2( g)

\(\Rightarrow\) m_{dd AgNO3}= \(\dfrac{10,2}{50\%}\)= 20,4( g)

ta có: m_{dd sau pư}= m_{dd CaCl2}+ m_{dd AgNO3}- m_AgCl

= 7,4+ 20,4- 0,06. 143,5

= 19,19( g)

có: m_Ca(NO3)2= 0,03.164= 4,92( g)

\(\Rightarrow\) C%_Ca(NO3)2= \(\dfrac{4,92}{19,19}\). 100%\(\approx\) 25,64%

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

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

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

Hỗn hợp thu được gồm có C a C O 3 , NaCl và C a C l 2  dư

Khi cho C O 2  ( n C O 2  = 6,70. 10 - 2 ) vào hỗn hợp, xảy ra phản ứng:

C O 2  +  C a C O 3  + H 2 O → C a ( H C O ₃ ) 2  (2)

Theo (2), số mol  C a C O 3  bị hòa tan = số mol  C O 2  phản ứng = 4,02. 10 - 2  (mol)

Khối lượng kết tủa  C a C O 3  thu được là:

(5,00. 10 - 2  - 4,02. 10 - 2 ) x 100 = 0,98 (g).

\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

a) Pt : \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl|\)

               1              2                  1                 2

             0,25          0,5               0,25 

b) \(n_{Fe\left(OH\right)2}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)

⇒ \(m_{Fe\left(OH\right)2}=0,25.90=22,5\left(g\right)\)

c) \(n_{FeCl2}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)

100ml = 0,1l

\(C_{M_{ddFeCl2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

d) Pt : \(Fe\left(OH\right)_2\underrightarrow{t^o}FeO+H_2O|\)

                1               1          1

               0,25          0,25

\(n_{FeO}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

⇒ \(m_{FeO}=0,25.72=18\left(g\right)\)

 Chúc bạn học tốt

a)
$BaCl_2 + Na_2SO_4 \to BaSO_4 + 2NaCl$

$n_{BaCl_2} = 0,01 = n_{Na_2SO_4} = 0,01 \Rightarrow $ Vừa đủ

$n_{BaSO_4} = n_{Na_2SO_4} = 0,01(mol)$
$m_{BaSO_4} = 0,01.233 = 0,233(gam)$
b)

$n_{NaCl} = 2n_{Na_2SO_4} = 0,02(mol)$
$V_{dd} = 0,1 + 0,2 = 0,3(lít)$
$C_{M_{NaCl}} = \dfrac{0,02}{0,3} = 0,067M$

c)

$[Na^+] = [Cl^-] = C_{M_{NaCl}} = 0,067M$

\(n_{BaCl_2}=0.1\cdot0.1=0.01\left(mol\right)\)

\(n_{Na_2SO_4}=0.2\cdot0.05=0.01\left(mol\right)\)

\(BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl\)

\(0.01..........0.01............0.01..............0.02\)

\(m_{BaSO_4}=0.01\cdot233=2.33\left(g\right)\)

\(C_{M_{NaCl}}=\dfrac{0.01}{0.1+0.2}=0.03\left(M\right)\)

\(\left[Na^+\right]=\left[Cl^-\right]=0.03\left(M\right)\)

bn tk lời giải nek:

https://youtu.be/TWOuSaUikOk