Ta có: \(n_{Ca\left(OH\right)_2}=\dfrac{m_{Ca\left(OH\right)_2}}{M_{Ca\left(OH\right)_2}}=\dfrac{120}{74}=\dfrac{60}{37}\left(mol\right)\)
a)Ta có:
\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
60/37 120/37 60/37
Ta có: \(mCaCl_2=nCaCl_2.MCaCl_2=\dfrac{60}{7}.111=180\left(g\right)\)
Đổi 500ml=0,5l
\(C_{M_{HCl}}=\dfrac{120}{37}:0,5=\dfrac{240}{37}M\)