PT ion: \(H^++OH^-\rightarrow H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{H^+}=n_{HCl}=0,05\cdot1,2=0,06\left(mol\right)\\n_{OH^-}=n_{KOH}=0,05\cdot1=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) OH^- còn dư 0,01 mol

\(\Rightarrow\left[OH^-\right]=\dfrac{0,01}{0,1}=0,1\left(M\right)\) \(\Rightarrow pH=14+log\left(0,1\right)=13\)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{FeO}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+72b=15,6\)  (2)

Bảo toàn electron: \(3n_{Fe}+n_{FeO}=3n_{NO}\) \(\Rightarrow3a+b=0,45\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,105\\b=0,135\end{matrix}\right.\)

Bài 3:

a) Ta có: \(\left[H^+\right]=0,005\cdot2=0,01\left(M\right)\) \(\Rightarrow pH=-log\left(0,01\right)=2\)

b) Ta có: \(\left[OH^-\right]=0,0005\cdot2=10^{-3}\left(M\right)\) \(\Rightarrow pH=14+log\left(10^{-3}\right)=11\)

c) Ta có: \(n_{H^+}=0,05\cdot0,1\cdot2+0,05\cdot0,05\cdot2=0,015\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{0,015}{0,1}=0,15\left(M\right)\) \(\Rightarrow pH=-log\left(0,15\right)\approx0,82\)

\(\overset{0}{Al}\rightarrow\overset{+3}{Al}+3e\)

\(\overset{+5}{2N}+8e\rightarrow\overset{+1}{N_2O}\)

PTHH:

\(8Al+30HNO_3\rightarrow8Al\left(NO_3\right)_3+3N_2O\uparrow+15H_2O\)

chả bt 

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Chọn D

\(PbO+CO \xrightarrow{t^{o}} Pb+CO_2\\ n_{CO_2}=0,3(mol)\\ n_{PbO}=n_{CO_2}=0,3(mol)\\ m_{PbO}=223.0,3=66,9(g)\)