a)
Ca + 2H2O → Ca(OH)2 + H2↑ (1)
Ca(OH)2 + 2HCl → CaCl2 + 2H2O (2)
nHCl = 0,5 . 0,2 = 0,1 mol
Theo (2): nCa(OH)2 =\(\frac{1}{2}\)nHCl = 0,05 mol
Theo (1): nCa = nCa(OH)2 = 0,05 mol
mCa = 0,05 . 40 = 2 (g)
b)
Theo (1): nH2 = nCa = 0,05 mol
mdd sau p.ứ (1) = mCa + mH2O - mH2
= 2 + 200 - 0,05 . 2
= 201,9 (g)
\(C\%_{Ca\left(OH\right)2}=\frac{0,05.74}{201,9}.100\%=1,83\%\)