a, \(n_{KOH}=0,3.1=0,3\left(mol\right)\)

PT: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

Theo PT: \(n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,15\left(mol\right)\)

\(\Rightarrow m_{K_2SO_4}=0,15.174=26,1\left(g\right)\)

b, \(C_{M_{K_2SO_4}}=\dfrac{0,3}{0,3+0,2}=0,6\left(M\right)\)

Câu 1:

Ta có: \(n_{H_3PO_4}=0,15.1=0,15\left(mol\right)\)

\(n_{NaOH}=0,225.1=0,225\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{H_3PO_4}}=\dfrac{0,225}{0,15}=1,5\)

⇒ Pư tạo muối NaH₂PO₄ và Na₂HPO₄.

a, Giả sử: \(\left\{{}\begin{matrix}n_{NaH_2PO_4}=x\left(mol\right)\\n_{Na_2HPO_4}=y\left(mol\right)\end{matrix}\right.\)

BTNT Na, có: x + 2y = 0,225 (1)

BTNT P, có: x + y = 0,15 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,075\left(mol\right)\\y=0,075\left(mol\right)\end{matrix}\right.\)

⇒ m_NaH2PO4 = 0,075.120 = 9 (g)

m_Na2HPO4 = 0,075.142 = 10,65 (g)

b, \(C_{M_{NaH_2PO_4}}=C_{M_{Na_2HPO_4}}=\dfrac{0,075}{0,375}=0,2M\)

Câu 2: 

Ta có: \(n_{H_3PO_4}=0,15\left(mol\right)\)

\(n_{KOH}=0,35\left(mol\right)\)

\(\Rightarrow\dfrac{n_{KOH}}{n_{H_3PO_4}}=2,333\)

Vậy: Pư tạo muối Na₂HPO₄ và Na₃PO₄.

a, Giả sử: \(\left\{{}\begin{matrix}n_{Na_2HPO_4}=x\left(mol\right)\\n_{Na_3PO_4}=y\left(mol\right)\end{matrix}\right.\)

BTNT Na, có: 2x + 3y = 0,35 (1)

BTNT P, có: x + y = 0,15 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{Na_2HPO_4}=0,1.142=14,2\left(g\right)\)

\(m_{Na_3PO_4}=0,05.164=8,2\left(g\right)\)

b, Ta có: \(\left\{{}\begin{matrix}C_{M_{Na_2HPO_4}}=\dfrac{0,1}{0,5}=0,2M\\C_{M_{Na_3PO_4}}=\dfrac{0,05}{0,5}=0,1M\end{matrix}\right.\)

Bạn tham khảo nhé!

Có: \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)

PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

______0,2_____0,1_______0,1 (mol)

a, \(V_{ddH_2SO_4}=\dfrac{0,1}{2}=0,05\left(l\right)\)

b, \(C_{M_{Na_2SO_4}}=\dfrac{0,1}{0,2+0,05}=0,4M\)

Bạn tham khảo nhé!

PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

            \(CuCl_2+2KOH\rightarrow2KCl+Cu\left(OH\right)_2\downarrow\)

a+b) Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(\Rightarrow n_{HCl}=0,2\left(mol\right)=n_{KOH}\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2\cdot36,5}{300}\cdot100\%\approx2,43\%\\C_{M_{KOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)

c) PTHH: \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

Theo các PTHH: \(n_{CuO\left(lý.thuyết\right)}=n_{Cu\left(OH\right)_2}=n_{Cu}=0,1\left(mol\right)\)

\(\Rightarrow n_{CuO}=0,1\cdot95\%=0,095\left(mol\right)\) \(\Rightarrow m_{CuO}=0,095\cdot80=7,6\left(g\right)\)