Answer
0,855 g AlCl$_3$
0,346 g H2O
Work Step by Step
First, we will find the molar mass of AlCl$_3$:
MM(AlCl3) = 78g/mol
In 0,500 g of Al(OH)$_3$ there are:
$\frac{0,500 g}{78g/mol} = 6,41\times10^{-3} mol AlCl_3$
$6,41\times10^{-3} mol AlCl_3$ reacts with HCl
The molar mass of HCl is 133,35 g/mol.
The mass of $6,41\times10^{-3}$ mols=$6,41\times10^{-3}$ mols x 133,35 g/mol = 0,855 g HCl
Second, we will find mass of water:
$6,41\times10^{-3} mol AlCl_3$ reacts with 3H$_2$O:
$6,41\times10^{-3} \times3$= 0,0192 mols.
The molar mass of H$_2$O is 18 g/mol
The mass of 0,0192 mols of water is:
0,0192 x 18 = 0,346 g H$_2$O
