Bài 3:
\(a)\) \(PTHH:\)
\(2K+2H_2O--->2KOH+H_2\) \((1)\)
\(K_2O+H_2O--->2KOH\) \((2)\)
\(b)\)
\(nH_2=\dfrac{5,6}{22,4}=0,25(mol)\)
Theo PTHH (1) \(nK=2.nH_2=2.0,25=0,5(mol)\)
\(=> mK=0,5.39=19,5(g)\)
\(=> mK_2O=58,5-19,5=39(g)\)
\(=> nK_2O=\dfrac{39}{94}=0,4(mol)\)
\(\%mK=\dfrac{19,5.100}{58,5}=33,33\%\)
\(=>\%mK_2O=100\%-33,33\%=66,67\%\)
\(c)\)
Theo (1) và (2) \(nKOH=0,5+0,8=1,3(mol)\)
\(=>mKOH=1,3.56= 72,8(g)\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
\(K_2O+H_2O\rightarrow2KOH\)
b)\(n_{H_2}:\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
2............2..................2............1(mol)
0,5.........0,5..............0,5..........0,25(mol)
\(m_K:0,5.39=19,5\left(g\right)\)
% Khối lượng K trong hh:\(\dfrac{19,5}{58,5}.100\%=33,3\%\)
% Khối lượng KOH trong hh: 100%-33,3%=66,6%
c) \(n_{K_2O}:\dfrac{58,5-19,5}{94}=\dfrac{39}{94}\left(mol\right)\)
\(K_2O+H_2O\rightarrow2KOH\)
1...............................2(mol)
\(\dfrac{39}{94}.................\dfrac{39}{47}\left(mol\right)\)
Khối lượng KOH:(0,5+\(\dfrac{39}{47}\)).56=74,5(g)
* Chắc thế này -_-*
