PTHH: CaO + H2O -> Ca(OH)2
Ta có: \(m_{dd}=11,2+188,8=200\left(g\right)\\ n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ =>n_{Ca\left(OH\right)_2}=n_{CaO}=0,2\left(mol\right)\\ =>m_{Ca\left(OH\right)_2}=0,2.74=14,8\left(g\right)\\ =>C\%_{ddCa\left(OH\right)_2}=\dfrac{14,8}{200}.100=7,4\%\)
PTHH: \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
---------0,2-----------------0,2 ( mol )
Ta có : \(n_{CaO}=\dfrac{11,2}{56}=0,2mol\)
theo PTHH => nCa(OH)2= 0,2 mol
=> \(m_{Ca\left(OH\right)_2}=0,2.74=14,8gam\)
khối lượng dung dịch : 11,2 + 188,8 = 200 gam
\(C\%=\dfrac{14,8}{200}.100\%=7,4\%4\)
vậy...
Ta có:\(m_{ddsauphảnứng}=11,2+188,8=200\left(g\right)\\ n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:CaO+H_2O->Ca\left(OH\right)_2\\ =>n_{Ca\left(OH\right)_2}=n_{CaO}=0,2\left(mol\right)\\ =>m_{Ca\left(OH\right)_2}=74.0,2=14,8\left(g\right)\\ =>C\%_{ddsauphảnứng}=\dfrac{14,8}{200}.100=7,4\%\)
