\(n_{NaCl}=\dfrac{1,17}{58,5}=0,02\left(mol\right)\\ n_{BaCl_2}=\dfrac{2,08}{208}=0,01\left(mol\right)\\ \left[Na^+\right]=\dfrac{0,02}{0,1}=0,2\left(M\right)\\ \left[Ba^{2+}\right]=\dfrac{0,01}{0,1}=0,1\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,02+0,01.2}{0,1}=0,4\left(M\right)\)

m_bacl2=150.10,4/100=15,6g=>n_bacl2=15,6/208=0,075mol

m_h2so4=250.9,8/100=24,5g=>n_h2so4=24,5/98=0,25mol

pt: bacl2 + h2so4 -> baso4 + 2 hcl

ncó_{: 0,075 0,25}

n_{pứ: 0,075 0,075 0,075 0,15}

n_{dư: 0 0,175}

dd x sau pứ hcl và h2so4 dư

=> m_baso4=0,075.233=17,475g

m_{dd x}=m_{dd bacl2} +m_{dd h2so4}-m_{baso4=150+250-17,475=382,525g}

=>m_hcl=0,175.36,5=6,3875g=>C%=6,3875/382,525.100~1,670%

=>m_{h2so4 dư}=0,175.98=17,15g=>C%=17,15/382,525.100~4,483%

⇒ M = 244g/mol = M B a C l 2 . x H 2 O . Từ đó :

⇒ x = 2.

Đáp số :  B a C l 2 . x H 2 O

Bài 1:

Ta có: \(n_{Fe}=0,1\left(mol\right)\)

PT: \(Fe+4HNO_3\underrightarrow{t^o}Fe\left(NO_3\right)_3+NO+2H_2O\)

___0,1_____0,4_____0,1_______0,1 (mol)

\(\Rightarrow m_{HNO_3}=0,4.63=25,2\left(g\right)\)

\(\Rightarrow m_{ddHNO_3}=\dfrac{25,2}{6,3\%}=400\left(g\right)\)

Ta có: m _{dd sau pư} = m_Fe + m _{dd HNO3} - m_NO = 5,6 + 400 - 0,1.30 = 402,6 (g)

\(\Rightarrow C\%_{Fe\left(NO_3\right)_3}=\dfrac{0,1.242}{402,6}.100\%\approx6,01\%\)

Bạn tham khảo nhé!

Bài 2 : 

n KMnO4 = 0,2(mol)

$Mn^{+7} + 5e \to Mn^{+2}$
$Fe^{+2} \to Fe^{+3} + 1e$

Bảo toàn electron  :

n FeSO4 = 5n KMnO4 = 0,2.5 = 1(mol)

n FeSO4.7H2O = n FeSO4 = 1(mol)

=> a = 1.278 = 278(gam)

Bài 1 : 

n Fe = 5,6/56 = 0,1(mol)

Bảo toàn electron :

3n Fe = 3n NO

=> n NO = 0,1(mol)

n HNO3 = 4n NO = 0,4(mol)

=> m HNO3 = 0,4.63 = 25,2(gam)

=> m dd HNO3 = 25,2/6,3% = 400(gam)

Sau phản ứng : 

n Fe(NO3)3 = n Fe = 0,1(mol)

m dd = 5,6 + 400 - 0,1.30 = 402,6(gam)

C% Fe(NO3)3 = 0,1.242/402,6   .100% = 6,01%

\(m_{BaCl_2}=\dfrac{200.10,4}{100}=20,8\left(g\right)\)

\(n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

\(a.BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

mol 0,1 → 0,1 0,1 0,2

b.\(m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

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

a. PTHH :

BaCl2 + H2SO4 ---> BaSO4 + 2HCl (1)

=> Kết tủa a là BaSO4 , b là dd HCl

b. ADCT : \(m_{ct}\) = \(\dfrac{m_{dd}\cdot C\%}{100\%}\)=> \(m_{BaCl_2}\)= \(\dfrac{200\cdot10,4\%}{100\%}\)= 20,8 (g)

ADCT : n = \(\dfrac{m}{M}\)=> \(n_{BaCl_2}\)= \(\dfrac{20,8}{208}\)= 0,1 (mol)

Theo (1) : \(n_{BaCl_2}\)= \(n_{BaSo_4}\) = 0,1 (mol)

ADCT : m=n.M => \(m_{BaSO_4}\)= 0,1.233= 23,3 (g)

ADĐLBTKL ta có :

\(m_{ddBaCl_2}\) + \(m_{ddH_2SO_4}\) = \(m_{BaSO_4}\) + \(m_{ddHCl}\)

=> \(m_{ddHCl}=\) 200 + 328,3 - 23,3 = 505 (g)

c. Ta có : \(V_{ddHCl}=\)505 : 1,01=500(ml)=0,5(l)

Theo (1) : \(n_{HCl}=2n_{BaCl_2}\)= 0,2 (mol)

ADCT : \(C_M\) = \(\dfrac{n}{V}\)=> \(C_{M_b}=\dfrac{0,2}{0,5}\)= 0,4M

d. ADCT : n = \(\dfrac{m}{M}\)=> \(n_{Fe}=\dfrac{4,2}{56}=0,075\left(mol\right)\)

PTHH:

Fe + 2HCl --> \(FeCl_2\) + \(H_2\)

mol trước pứ : 0,075 0,2 (mol)

mol trong pứ: 0,075->0,15--------------->0,075 (mol)

mol sau pứ : 0 -->0,05---------------->0,075 (mol)

=> Fe hết, HCl dư

ADCT : V=n.22,4=> \(V_{H_2}\)= 0,075.22,4=1,68 (l)

\(m_{CuSO_4}=\dfrac{200\cdot32}{100}=64\left(g\right)\)\(\Rightarrow n_{CuSO_4}=\dfrac{64}{160}=0,4mol\)

\(m_{BaCl_2}=\dfrac{200\cdot10,4}{100}=20,8\left(g\right)\)\(\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1mol\)

\(CuSO_4+BaCl_2\rightarrow BaSO_4\downarrow+CuCl_2\)

0,4             0,1           0,1             0,1

b)\(m_{BaSO_4}=0,1\cdot233=23,3\left(g\right)\)

c)\(m_{CuCl_2}=0,1\cdot135=13,5\left(g\right)\)

 \(\Rightarrow m_{ddsau}=200+200-13,5=386,5\left(g\right)\)

 \(\Rightarrow C\%=\dfrac{23,3}{386,5}\cdot100\%=6,028\%\)