Bạn ơi , câu 1 sao dd NaCl lại tác dụng với NaCl v ?

Bài 1:
\(m_{H_2SO_4.40\%}=200\times40\%=80\left(g\right)\)

\(\Rightarrow m_{H_2O}=200-80=120\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4.19,6\%}=\frac{120}{100\%-19,6\%}=149,25\left(g\right)\)

\(\Rightarrow m_{H_2SO_4.19,6\%}=149,25\times19,6\%=29,253\left(g\right)\)

\(\Rightarrow m_{H_2SO_4}=80-29,253=50,747\left(g\right)\)

Bài 2:

Gọi \(m_{ddNaCl.20\%}=x\left(g\right)\Rightarrow m_{NaCl.20\%}=20\%x=0,2x\left(g\right)\)

\(m_{ddNaCl.30\%}=y\left(g\right)\Rightarrow m_{NaCl.30\%}=30\%y=0,3y\left(g\right)\)

\(m_{NaCl.26\%}=300\times26\%=78\left(g\right)\)

Ta có hệ: \(\left\{{}\begin{matrix}x+y=300\\0,2x+0,3y=78\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=120\\y=180\end{matrix}\right.\)

Vậy \(m_{ddNaCl.20\%}=120\left(g\right)\)

\(m_{ddNaCl.30\%}=180\left(g\right)\)

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)

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

PTHH: \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_4\downarrow\) (1)

a) Ta có: \(\left\{{}\begin{matrix}m_{BaCl_2}=150\cdot5,2\%=7,8\left(g\right)\\m_{H_2SO_4}=250\cdot19,6\%=49\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{BaCl_2}=\frac{7,8}{208}=0,0375\left(mol\right)\\n_{H_2SO_4}=\frac{49}{98}=0,5\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\frac{0,0375}{1}< \frac{0,5}{1}\) \(\Rightarrow\) BaCl₂ phản ứng hết, H₂SO₄ còn dư

\(\Rightarrow n_{BaSO_4}=0,0375mol\) \(\Rightarrow m_{BaSO_4}=0,0375\cdot233=8,7375\left(g\right)\)

b) Dung dịch A gồm: \(HCl\) và \(H_2SO_{4\left(dư\right)}\)

PTHH: \(HCl+NaOH\rightarrow NaCl+H_2O\) (2)

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\) (3)

Theo PTHH (1): \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{BaCl_2}=0,075mol=n_{HCl\left(2\right)}\\n_{H_2SO_4\left(dư\right)}=0,4625mol=n_{H_2SO_4\left(3\right)}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH\left(2\right)}=0,075mol\\n_{NaOH\left(3\right)}=0,925mol\end{matrix}\right.\) \(\Rightarrow n_{NaOH}=1mol\)

\(\Rightarrow V_{NaOH}=\frac{1}{1,5}\approx0,67\left(l\right)=670\left(ml\right)\)

n_BaCl2= 0,1 (mol)

n_H2SO4 = 0,2327 (mol)

BaCl₂ + H₂SO₄ \(\rightarrow\) BaSO₄ \(\downarrow\) + 2HCl

bđ 0,1 0,2327 }

pư 0,1 \(\rightarrow\) 0,1 \(\rightarrow\) 0,1 \(\rightarrow\) 0,2 } (mol)

spư 0 0,1327 0,1 0,2 }

m_BaSO4 = 0,1 . 233 = 23,3 (g)

mdd(sau pư)= 400 + 1,14 . 100 - 23,3 =490,7 (g)

C%_(H2SO4)=\(\frac{0,1327.98}{490,7}\) . 100% = 2,65%

C% (HCl) =\(\frac{0,2.36,5}{490,7}\) . 100% = 1,49%

200ml = 0,2l

300ml = 0,3l

\(n_{CuCl2}=3.0,3=0,9\left(mol\right)\)

a) Pt : \(2NaOH+CuCl_2\rightarrow2NaCl+Cu\left(OH\right)_2|\)

                 2              1                2               1

                               0,9             1,8             0,9

    A : Là natri clorua 

    B : đồng (II) hidroxit 

b) \(n_{Cu\left(OH\right)2}=\dfrac{0,9.1}{1}=0,9\left(mol\right)\)

⇒ \(m_{Cu\left(OH\right)2}=0,9.98=88,2\left(g\right)\)

c) \(n_{NaCl}=\dfrac{0,9.2}{1}=1,8\left(mol\right)\)

\(V_{ddspu}=0,2+0,3=0,5\left(l\right)\)

\(C_{M_{NaCl}}=\dfrac{1,8}{0,5}=3,6\left(M\right)\)

 Chúc bạn học tốt

mBaCl2 = 400.5,2% = 20,8g 
mH2SO4 = 100.1,14.20% = 22,8g 

BaCl2 + H2SO4 = BaSO4 + 2HCl 
208...... 98........... 233........ 2.36,5g 
20,8..... 22,8......... 

20,8/208 < 22,8/98 => BaCl2 phản ứng hết và H2SO4 còn dư 
=> mBaSO4 = 20,8.233/208 = 23,3g 
=> mHCl = 20,8.2.36,5/208 = 7,3g 
=> mH2SO4 (dư) = 22,8 - 20,8.98/208 = 13g 

mdd (trước ph.ư) = 400 + 100.1,14 = 514g 
mdd (sau ph.ư) = 514 - 23,3 = 490,7g 

=> C%HCl = 7,3/490,7 = 1,49% 
=> C%H2SO4 (dư) = 13/490,7 = 2,65% 

nbacl2= 400*5,2/(100*208)=0,1 mol

mdd h2so4= 100*1,14=114g

nh2so4=114*20/(100*98)=57/245 mol

bacl2+ h2so4-> baso4+ 2hcl

thấy 0,1< 57/245

=> sau pư bacl2 hết, h2so4 còn dư

nh2so4 dư=57/245-0,1=13/98

n baso4= nbacl2= 0,1 mol

=> mbaso4= 0,1*233=23,3g

mdd sau pư=400+114-23,3=490,7g

C% h2so4 dư=13/98*98*100/490,7=2,65%

C% hcl=0,1*2*36,5*100/490,7=1,49%