a)
\(n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\\ PTHH:K_2O+H_2O\rightarrow2KOH\\ n_{KOH}=2.0,4=0,8\left(mol\right)\\ m_{KOH}=0,8.56=44,8\left(g\right)\\ C\%_{ddKOH}=\dfrac{44,8}{500}.100=8,96\%\\ \)
b)
\(n_{CuCl_2}=0,25.2=0,5\left(mol\right)\\ PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\)
Ta có: \(\dfrac{0,5}{1}>\dfrac{0,8}{2}\)
-> KOH hết, CuCl2 dư => Tính theo nKOH.
=> \(n_{Cu\left(OH\right)_2}=\dfrac{n_{KOH}}{2}=\dfrac{0,8}{2}=0,4\left(mol\right)\\ m_{kt}=m_{Cu\left(OH\right)_2}=0,4.98=39,2\left(g\right)\)
nK2O = 37.6/94 = 0.4 (mol)
K2O + H2O => 2KOH
0.4.......................0.8
mKOH = 0.8*56 = 44.8 (g)
C% KOH = 44.8/500 * 100% = 8.96%
b)
nCuCl2 = 0.25*2=0.5 (mol)
CuCl2 + 2KOH => Cu(OH)2 + 2KCl
LTL : 0.5/1 > 0.8/2 => CuCl2 dư
nCu(OH)2 = 0.4 (mol)
mCu(OH)2 = 0.4*98 39.2 (g)
