\(n_{HCl}=0.2\cdot1.5=0.3\left(mol\right)\)

\(n_{KOH}=0.15\cdot2=0.3\left(mol\right)\)

\(KOH+HCl\rightarrow KCl+H_2O\)

\(0.3..........0.3..........0.3\)

\(m_{KCl}=0.3\cdot74.5=22.35\left(g\right)\)

\(C_{M_{KCl}}=\dfrac{0.3}{0.2+0.15}=0.85\left(M\right)\)

3.

\(n_{Ba^{2+}}=0,5.0,2=0,1\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{0,1}{0,2+0,4}=0,17M\)

\(n_{Cl^-}=2.0,5.0,2=0,2\left(mol\right)\Rightarrow\left[Cl^-\right]=\dfrac{0,2}{0,2+0,4}=0,33M\)

\(n_{Na^+}=2.0,2.0,4=0,16\left(mol\right)\Rightarrow\left[Na^+\right]=\dfrac{0,16}{0,2+0,4}=0,27M\)

\(n_{SO_4^{2-}}=0,2.0,4=0,08\left(mol\right)\Rightarrow\left[SO_4^{2-}\right]=\dfrac{0,08}{0,2+0,4}=0,13M\)

4.

\(n_{H^+}=n_{Cl^-}=2.0,15=0,3\left(mol\right)\Rightarrow\left[Cl^-\right]=\left[H^+\right]=\dfrac{0,3}{0,15+0,05}=1,5M\)

\(n_{Ba^{2+}}=0,05.2,8=0,14\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{0,14}{0,15+0,05}=0,7M\)

\(n_{OH^-}=2.0,05.2,8=0,28\left(mol\right)\Rightarrow\left[OH^-\right]=\dfrac{0,28}{0,15+0,05}=1,4M\)

nHCl=0,15 . 2=0,3(mol)

PTHH: CuO + 2 HCl -> CuCl2 + H2O

x___________2x____x(mol)

ZnO +2 HCl -> ZnCl2 + H2O

y______2y_____y(mol)

Ta có hpt: \(\left\{{}\begin{matrix}80x+81y=12,1\\2x+2y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

Vddsau=VddHCl=0,15(l)

=> CMddCuO= 0,05/0,15=1/3(M)

CMddZnO=0,1/0,15=2/3(M)

a, \(CuO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)

b, \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{CuO}=0,2\left(mol\right)\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

c, \(n_{\left(CH_3COO\right)_2Cu}=n_{Cu}=0,1\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Cu}=0,1.182=18,2\left(g\right)\)

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, \(m_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

Theo PT: \(n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow m_{MgCl_2}=0,3.95=28,5\left(g\right)\)

c, \(n_{HCl}=2n_{Mg}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)

\(V_{ddKCl}=\dfrac{0,15}{2}=0,075\left(l\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2mol\)

\(n_{H_2SO_4}=0,04.1=0,04mol\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

0,2  >   0,04                              ( mol )

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

\(m_{ZnSO_4}=0,04.161=6,44g\)

Câu b ko hiểu lắm bạn ơi!

100ml = 0,1l

150ml = 0,15l

\(n_{NaOH}=1.0,1=0,1\left(mol\right)\)

\(n_{CuSO4}=1,2.0,15=0,18\left(mol\right)\)

a) Pt : \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2|\)

                2                1                1                   1 

              0,1             0,18           0,05              0,05

b) Hiện tượng : màu xanh lam của dung dịch CuSO₄ nhạt dần , xuất hiện kết tủa có màu xanh lơ

c) Lập tỉ số so sánh : \(\dfrac{0,1}{2}< \dfrac{0,18}{1}\)

                  ⇒ NaOH phản ứng hết , CuSO₄ dư

                  ⇒ Tính toán dựa vào số mol của NaOH

\(n_{Cu\left(OH\right)2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)

⇒ \(m_{cu\left(OH\right)2}=0,05.98=4,9\left(g\right)\)

d) \(n_{Na2SO4}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

\(n_{CuSO4\left(dư\right)}=0,18-\left(\dfrac{0,1.1}{2}\right)=0,13\left(mol\right)\)

\(V_{ddspu}=0,1+0,15=0,25\left(l\right)\)

\(C_{M_{Na2SO4}}=\dfrac{0,05}{0,25}=0,2\left(M\right)\)

\(C_{M_{CuSO4\left(dư\right)}}=\dfrac{0,13}{0,25}=0,52\left(M\right)\)

 Chúc bạn học tốt

a. 2NaOH + CuSO₄ ---> Cu(OH)₂ + Na₂SO₄.

a: \(n_{H_2SO_4}=0.25\cdot2=0.5\left(mol\right)\)

\(n_{Al_2O_3}=\dfrac{10.2}{27\cdot2+16\cdot3}=0.1\left(mol\right)\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

0,1              0,5

Vì 0,1/1<0,5/3

nên Al2O3 hết, H2SO4 dư

=>Tính theo Al2O3

b: 

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

0,1               0,3           0,1

\(m_{H_2SO_4\left(pư\right)}=0.3\cdot98=29.4\left(g\right)\)

\(m_{muối}=0.1\left(54+3\cdot96\right)=34.2\left(g\right)\)