a) Ta có: \(\left\{{}\begin{matrix}\left[Cu^{2+}\right]=C_{M_{Cu\left(NO_3\right)_2}}=0,3\left(M\right)\\\left[NO_3^-\right]=2C_{M_{Cu\left(NO_3\right)_2}}=0,6\left(M\right)\end{matrix}\right.\)

b) Ta có: \(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\) \(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=0,5\left(M\right)\\\left[SO_4^{2-}\right]=0,25\left(M\right)\end{matrix}\right.\)

a,\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

PTHH: Na₂O + H₂O → 2NaOH

Mol:     0,25                      0,5

\(C_{M_{ddNaOH}}=\dfrac{0,5}{0,5}=1M\)

b, PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

    Mol:       0,5           0,25          0,25

\(m_{ddH_2SO_4}=\dfrac{0,2.98.100\%}{20\%}=98\left(g\right)\)

\(V_{ddH_2SO_4}=\dfrac{98}{1,14}=85,96\left(ml\right)\)

c,V_{dd sau pứ} = 0,5 + 0,08596 = 0,58596 (l)

 \(C_M=\dfrac{0,25}{0,58596}=0,427M\)

\(a.C_M=\dfrac{0,06}{1,5}=0,04M\\ b.C_M=\dfrac{\dfrac{400}{160}}{4}=0,625M\\ c.C_M=\dfrac{\dfrac{10,53}{58,5}}{\dfrac{450}{1,25}:1000}=0,5M\\ d.C_M=\dfrac{\dfrac{70,2}{40}}{0,5}=3,51M\\ e.C_M=\dfrac{\dfrac{42}{200}}{\dfrac{742}{1,3}:1000}=0,368M\)

a)

nFe2O3=16/160=0,1(mol)

nHCl=0,5.1=0,5(mol)

PTHH: Fe2O3 + 6 HCl -> 2 FeCl3 + 3 H2O

Ta có: 0,1/1 > 0,5/6

=> HCl hết, Fe2O3 dư, tính theo nHCl.

nFeCl3= 2/6. nHCl= 2/6 . 0,5= 1/6(mol)

=>mFeCl3= 162,5. 1/6= 27,083(g)

b) Vddsau=VddHCl=0,5(l)

- dd sau p.ứ chỉ có FeCl3.

=> CMddFeCl3= 1/6: 0,5= 1/3(M)

a, \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)=3360\left(ml\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{FeCl_2}}=\dfrac{0,15}{0,2}=0,75\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\end{matrix}\right.\)

c, Ta có: \(m_{ddHCl}=1,25.200=250\left(g\right)\)

⇒ m _{dd sau pư} = 8,4 + 250 - 0,15.2 = 258,1 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%\approx7,38\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{258.1}.100\%\approx1,41\%\end{matrix}\right.\)

a) \(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)

\(n_{NaOH}=\dfrac{8\%.500}{40}=1\left(mol\right)\)

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

0,5---------------->0,5------->0,25

\(\Sigma n_{NaOH}=0,5+1=1,5\left(mol\right)\)

\(m_{ddsaupu}=11,5+500-0,25.2=511\left(g\right)\)

=> \(C\%_{NaOH}=\dfrac{1,5.40}{511}.100=11,74\%\)

b) Gọi thể tích dung dịch X cần tìm là V

 \(n_{H^+}=V.1+V.0,5.1=2V\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

Ta có : \(n_{H^+}=n_{OH^-}=1,5\left(mol\right)\)

=> 2V=1,5

=> V=0,75(lít)