a. $n_{Ca(OH)2}$=$\frac{14,8}{74}$ =0,2(mol)
$Ca(OH)_{2}$+2$HCl$⇒$CaCl_{2}$+2$H_{2}O$
0,2 ..................0,4
⇒$m_{HCl}$=0,4×36,5=14,6(gam)
b. $m_{dd}$HCl=$\frac{14,6×100}{7,3}$ =200(g)
⇒$m_{dd}$$CaCl_{2}$=200+14,8=214,8(g)
$n_{CaCl_{2}}$=$n_{Ca(OH)2}$=0,2(mol)
⇒$m_{CaCl_{2}}$=0,2×111=22,2(g)
⇒$C%_{CaCl_{2}$=$\frac{22,2}{214,8}$ ×100%=10,34%
c. $n_{CaCl_{2}}$ =$\frac{27,75}{111}$= 0,25(mol)
⇒$n_{Ca(OH)2}$=$n_{CaCl_{2}}$=0,25(mol)
⇒$m_{Ca(OH)2}$=0,25×74=18,5(g)