C là \(BaSO_4\), D là \(HCl\)

\(a,PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ n_{BaCl_2}=\dfrac{31,2}{208}=0,15\left(mol\right)\\ \Rightarrow n_{BaSO_4}=0,15\left(mol\right)\\ \Rightarrow m_{BaSO_4}=0,15\cdot233=34,95\left(g\right)\\ b,n_{HCl}=2n_{BaCl_2}=0,3\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,3\cdot36,5=10,95\left(g\right)\\ m_{dd_{HCl}}=31,2+100-34,95=96,25\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{10,95}{96,25}\cdot100\%\approx11,38\%\)

a, \(H_2SO_4+BaCl_2\rightarrow BaSO_{4\downarrow}+2HCl\)

b, Ta có: \(m_{H_2SO_4}=114.20\%=22,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{22,8}{96}=0,2375\left(mol\right)\)

\(m_{BaCl_2}=400.5,2\%=20,8\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2375}{1}>\dfrac{0,1}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=n_{BaCl_2}=0,1\left(mol\right)\\n_{HCl}=2n_{BaCl_2}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,2375-0,1=0,1375\left(mol\right)\)

Ta có: m _{dd sau pư} = 114 + 400 - 23,3 = 490,7 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1375.98}{490,7}.100\%\approx2,75\%\\C\%_{HCl}=\dfrac{0,2.36,5}{490,7}.100\%\approx1,49\%\end{matrix}\right.\)

Bài 1

n_BaCl2= 200 *2.6%= 5.2 (g) ; n_BaCl2= 5.2/208=0.025(mol)

n_H2SO4=49*10%=4.9(g) ; n_H2SO4=4.9/98=0.05(mol)

PTHH

..........................H₂SO₄ + BaCl₂ ➞ 2HCl + BaSO₄

Trước phản ứng:0.05 : 0.025...................................(mol)

Trong phản ứng:0.025 : 0.025......... : 0.025 : 0.05(mol)

Sau phản ứng : 0.025 : 0 ......... : 0.025 : 0.05 (mol)

a) m_BaSO4=0.025*233=5.825(g)

b) m_{dd sau phản ứng} = 49+200-5.825=243.175(g)

C% (H₂SO₄) = (0.025* 98)/243.175*100%=1.007%

C% (HCl) = (0.05*36.5)/243.175*100%=0.007%

Bài 2:

n_HCl= 73 *25%= 18.25 (g) ; n_HCl= 18.25/36.5=0.5(mol)

n_AgNO3=34*5%=1.7(g) ; n_AgNO3=1.7/170=0.01(mol)

PTHH

..........................HCl + AgNO₃ ➞ AgCl + 2HNO₃

Trước phản ứng:0.5 : 0.01......................................(mol)

Trong phản ứng:0.01 : 0.01.............. : 0.01 : 0.01(mol)

Sau phản ứng : 0.49: 0 ............... : 0.01 : 0.01(mol)

a) m_AgCl=0.01*143.5=1.435(g)

b) m_{dd sau phản ứng} = 73+34-1.435=105.565(g)

C% (HNO₃) = (0.01* 63)/105.565*100%=0.0059%

C% (HCl) = (0.49*36.5)/105.565*100%=16.94%

a/ PTHH :      CuO + H2SO4 ===> CuSO4 + H2O

                      0,04         0,04                  0,04                 ( mol )

b/ m_H2SO4= 150 x 10% = 15 gam

=> n_H2SO4= 15 : 98 = 0,15 mol

n_CuO = 3,2 : 80 = 0,04 mol

Theo pt ta thấy CuO pứ hết, H2SO4 dư.

Ta lập tỉ lệ số mol theo pt:

=> m_{CuO pứ}= 3,2 gam

      m_CuSO4= 0,04 x 160 = 6,4 gam

c/ m_{dung dịch thu đc} = 3,2 + 150 = 153,2 gam

n_{H2SO4 dư}= 0,15 - 0,04 = 0,11 mol

=> m_{H2SO4 dư} = 0,11 x 98 = 10,78 gam

=> C%_H2SO4= 10,78 / 153,2 x 100% = 7,03%

      C%_CuSO4= 6,4 / 153,2 x 100% = 4,18%

\(n_{BaCl_2}=\dfrac{150.16,64\%}{137+35.2}=0,12\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.14,7\%}{98}=0,15\left(mol\right)\)

Phương trình hóa học : 

BaCl₂ + H₂SO₄ -----> BaSO₄ + 2HCl 

Dễ thấy \(\dfrac{n_{BaCl_2}}{1}< \dfrac{n_{H_2SO_4}}{1}\Rightarrow H_2SO_4\text{ dư }0,15-0,12=0,03\left(mol\right)\)

c) Khối lượng kết tủa : 

\(m_{BaSO_4}=0,12.233=27,96\) (g) 

Khối lượng chất tan : \(m_{HCl}=0,24.36,5=8,76\left(g\right)\) ; 

\(m_{H_2SO_4\left(\text{dư}\right)}=0,03.98=2,94\left(g\right)\) 

c) \(C\%_{H_2SO_4}\)= \(\dfrac{2,94}{150+100}.100\%=1,176\%\)

_{\(C\%_{HCl}=\dfrac{8,76}{150+100}.100\%=3.504\%\)}

d) NaOH + HCl ---> NaCl + H₂O 

      0,24 <-- 0,24

       mol       mol

    2NaOH + H₂SO₄  ---> Na₂SO₄ + 2H₂O

  0,06 mol <-- 0,03 mol  

\(\Rightarrow n_{NaOH}=0,24+0,06=0,3\left(mol\right)\)

\(V_{NaOH}=0,3.2=0,6\left(l\right)\)

a. PTHH    BaCl2 + H2SO4 → BaSO4↓ + 2HCl

b. _ mBaCl2 = 16,61.150:100 = 24,96 g

nBaCl2 = 24.96:208 = 0,12 mol

_ mH2SO4 = 14,7.100:100 = 14,7 g

nH2SO4 = 14,7:98 = 0,15 mol

Ta thấy: nBaCl2 < nH2SO4 ⇒ BaCl2 hết, H2SO4 dư

a) \(m_{HCl}=200.7,3\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl₂ + H₂O

Mol:      0,2        0,4          0,2

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

c, m_{dd sau pứ} = 16+200 = 216 (g)

\(C\%_{ddCuCl_2}=\dfrac{0,2.135.100\%}{216}=12,5\%\)

a)b)c)d) mBaCl2=150.16,64%=24,96g

=>nBaCl2=0,12 mol

mH2SO4=100.14,7%=14,7g=>nH2SO4=0,15mol

     BaCl2       + H2SO4 =>BaSO4    +2HCl

Bđ: 0,12 mol;    0,15 mol

Pứ: 0,12 mol=>0,12 mol=>0,12 mol=>0,24 mol

Dư:                   0,03 mol

Dd ban đầu chứa BaCl2 0,12 mol và H2SO4 0,15 mol

Dd A sau phản ứng chứa HCl 0,24 mol và H2SO4 dư 0,03 mol

mHCl=0,24.36,5=8,76g

mH2SO4=0,03.98=2,94g

Kết tủa B là BaSO4 0,12 mol=>mBaSO4=0,12.233=27,96g

mddA=mddBaCl2+mddH2SO4-mBaSO4

=150+100-27,96=222,04g

C%dd HCl=8,76/222,04.100%=3,945%

C% dd H2SO4=2,94/222,04.100%=1,324%

e) HCl     +NaOH =>NaCl +H2O

0,24 mol=>0,24 mol

H2SO4 +2NaOH =>Na2SO4 + 2H2O

0,03 mol=>0,06 mol

TÔNG nNaOH=0,3 mol

=>V dd NaOH=0,3/2=0,15 lit

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

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

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

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

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

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

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

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